Construction and Working Principle of Cyclotron Class 12 Physics

We will study here about the Construction and Working Principle of Cyclotron Class 12

  1. Working Principle of Cyclotron
  2. Construction of Cyclotron

Cyclotron Introduction

A cyclotron is used for accelerating positive ions, so that they acquire energy large enough to carry out nuclear reactions.

Cyclotron was designed by Lawrence and Livingstone in 1931.

In a cyclotron, the positive ions cross again and again the same alternating (radio frequency) electric field.

And gain the energy each time = q V.

q = charge and $V=p o t^{n}$ . difference in betn dees.

It is achieved by making them to move along spiral path under the action of a strong magnetic field.

Working Principle of Cyclotron :

A positive ion can acquire sufficiently large energy with a comparatively smaller alternating potential difference by making them to cross the same electric field again and again by making use of a strong magnetic field.

 

Construction of Cyclotron :

It consists of two D-shaped hollow semicircular metal chambers $\mathrm{D}_{1}$ and $\mathrm{D}_{2}$, called dees.

The two dees are placed horizontally with a small gap separating them.

The dees are connected to the source of high frequency electric field.

The dees are enclosed in a metal box containing a gas at a low pressure of the order of

10–3 mm mercury.

The whole apparatus is placed between the two poles of a strong electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.

The positive ions are produced in the gap between the two dees by the ionisation of the gas.

To produce proton, hydrogen gas is used; while for producing $\alpha$ -particles, helium gas is used. Theory : Consider that a positive ion is produced at the centre of the gap at the time, when the dee $D_{1}$ is at positive potential and the dee $D_{2}$, is at a negative potential. The positive ion will move from dee $D_{1}$ to dee $D_{2}$ The force on the positive ion due to magnetic field provides the centripetal force to the positive ion and it is deflected along a circular path because magnetic field is normal to the motion. Let strength of the magnetic field $=\mathrm{B}$ mass of ion $=\mathrm{m}, \quad$ velocity of ion $=\mathrm{v} \quad$ and $\quad$ charge of the positive ion $=\mathrm{q}$ and the radius of the semi-circular path $=\mathrm{r}$

then $\left.\quad \mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \quad \text { [inside the dee } \mathrm{D}_{2}\right]$

Thus, $r=\frac{m v}{B q}$

After moving along the semi-circular path inside the dee $D_{2},$ the positive ion reaches the gap between the dees.

At this stage, the polarity of the dees just reverses due to alternating “electric field” i.e. dee $D_{1}$, becomes negative and dee $\mathrm{D}_{2}$ becomes positive.The positive ion again gains the energy, as it is attracted by the dee $D_{1}$, After moving along the semi-circular path inside the dee $D_{1}$, the positive ion again reaches the gap and it gains the energy. ( $=\mathrm{q} \mathrm{V}$ ) This process repeats itself because, the positive ion spends the same time inside a dee irrespective of its velocity or the radius of the circular path.

The time spent inside a dee to cover semi-circular path,

is $\quad \mathrm{t}=\frac{\text { length of the semi circular path }}{\text { velocity }}=\frac{\pi \mathrm{r}}{\mathrm{v}}$

Or $\mathrm{t}=\frac{\pi \mathrm{m}}{\mathrm{Bq}} \quad\left[\frac{\mathrm{r}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{Bq}}\right]$

As positive ion gains kinetic energy its velocity increases, due to increasing velocity, decrease in time spent inside a dee of positive ions is exactly compensated by the increase in length of the semi circular path (r $\propto$ v).

Due to this condition, the positive ion always crosses the alternating electric field across the gap in correct phase.

 

              Also Read:

 

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Force Between Two Parallel Current Carrying Conductors || Class 12 Physics Notes

It is experimentally established fact that two current carrying conductors attract each other when the current is in same direction and repel each other when the current are in opposite direction. Here we will study about Force Between Two Parallel Current Carrying Conductors as wire:

Force Between parallel current carrying wires

Consider two long wires $W_{1}$ and $W_{2}$ kept parallel to each other and carrying currents $I_{1}$ and $\mathrm{I}_{2}$ respectively in the same direction. The separation between the wires is d. Consider a small element d\ell of the wire $WA_{2}$ The magnetic field at d\ell due to the wire $W_{1}$ is

$B_{1}=\frac{\mu_{0} I_{1}}{2 \pi d}$ …….(i)

The field due to the portions of the wire $W_{2},$ above and below $d \ell,$ is zero. Thus, eq” (i) gives

the net field at $\mathrm{d} \ell$ . The direction of this field is perpendicular to the plane of the diagram and going into it. The magnetic force at the element $d \ell$ due to wire $w_{1}$ is.

The vector product $\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$ has a direction towards the wire $\mathrm{W}_{1} .$ Thus, the length $\mathrm{d} \ell$ of wire $\mathrm{W}_{2}$ is attracted towards the wire $\mathrm{W}_{1}$. The force per unit length of the wire $\mathrm{W}_{2}$ due to the wire $W_{1}$ is

The vector product $\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$ has a direction towards the wire $W_{1}$ . Thus, the length $\mathrm{d} \ell$ of wire $\mathrm{W}_{2}$ is attracted towards the wire $\mathrm{W}_{1}$. The force per unit length of the wire $\mathrm{W}_{2}$ due to the wire $W_{1}$ is

If we take an element $\mathrm{d} \ell$ in the wire $\mathrm{W}_{1}$ and calculate the magnetic force per unit length on wire $W_{1}$ due to $\mathrm{W}_{2},$ it is again given by $\operatorname{eq}^{n}(i i)$

If the parallel wires currents in opposite directions, the wires repel each other.

The wires attract each other if current in the wires are flowing in the same direction.

And they repel each other if the currents are in opposite directions.

Experimental Demonstration

Definition of Ampere

$\because \quad \mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}} \mathrm{N} / \mathrm{m}$

When $\mathrm{I}_{1}=\mathrm{I}_{2}=1$ ampere and $\mathrm{r}=1 \mathrm{m},$ then $\quad \mathrm{F}=\frac{\mu_{0}}{2 \pi}=\frac{4 \pi \times 10^{-7}}{2 \pi} \mathrm{N} / \mathrm{m}=2 \times 10^{-7} \mathrm{N} / \mathrm{m}$ This leads to the following definition of ampere.

One ampere is that current which, if passed in each of two parallel conductors of infinite length and one metre apart in vacuum causes each conductor to experience a force of $2 \times 10^{-7}$ newton per metre of length of conductor.

Dimensional of formula of $\mu_{0}$

$\because \quad \mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}} \quad$ so

$\left[\mu_{0}\right]=\frac{[\mathrm{F}][\mathrm{r}]}{\left[\mathrm{I}_{1} \mathrm{I}_{2}\right]}=\frac{\left[\mathrm{ML}^{0} \mathrm{T}^{-2}\right][\mathrm{L}]}{\left[\mathrm{I}^{2}\right]}=\left[\mathrm{MLT}^{-2} \mathrm{I}^{-2}\right]$

 

Also Read:

Biot Savart’s Law

 

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Force on a Moving Charge in a Magnetic Field | Class 12 Physics Notes
  1. Force on a Charged Particle in a Magnetic Field
  2. Difference in Force on a Charged Particle by Magnetic Field and Electric Field

When a charged particle travels through a magnetic field, it experiences a force unlike any other that we’re familiar with in everyday life. To illustrate the point, envision yourself walking down the sidewalk, when all of a sudden, a strong gust of wind hits you from the side. Now imagine that instead of moving sideways, you shoot straight up to the sky. Here we will study about the Force on a Moving Charge in a Magnetic Field.

Force on a Charged Particle in a Magnetic Field

Force experienced by a current element Id $\vec{\ell}$ in magnetic field $\overrightarrow{\mathrm{B}}$ is given by

Now if the current element $\mathrm{Id} \vec{\ell}$ is due to the motion of charge particles, each particle having a charge q moving with velocity $\overrightarrow{\mathrm{v}}$ through a cross-section S,

$\mathrm{Id} \vec{\ell}=\mathrm{n} \mathrm{S} \mathrm{q} \quad \overrightarrow{\mathrm{v}} \cdot \mathrm{d} \ell=\mathrm{n} \mathrm{d} \tau \mathrm{q} \overrightarrow{\mathrm{v}}$ [with volume $\mathrm{d} \tau=\mathrm{S} \mathrm{d} \ell]$

$n \mathrm{d} \tau=$ the total number of charged particles in volume d\tau $(n=$ number of charged particles per unit volume),

force on a charged particle From this it is clear that : $\left.\vec{F}=\frac{1}{n} \frac{d \vec{F}}{d \tau}=q \quad \vec{v} \times \vec{B}\right)$

(a) The force $\overrightarrow{\mathrm{F}}$ is always perpendicular to both the velocity $\overrightarrow{\mathrm{v}}$ and the field $\overrightarrow{\mathrm{B}}$.

 

(b) A charged particle at rest in a steady magnetic field does not experience any force.

If the charged particle is at rest then $\overrightarrow{\mathrm{v}}=0,$ so $\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}=0$

 

(c) A moving charged particle does not experience any force in a magnetic field if its motion is parallel or antiparallel to the field.

i.e., if $\quad \theta=0^{\circ}$ or $180^{\circ}$

 

(d) If the particle is moving perpendicular to the field.

In this situation all the three vectors $\overrightarrow{\mathrm{F}}, \overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{B}}$ are mutually perpendicular to each other. Then $\sin \theta=\max =1,$ i.e., $\theta=90^{\circ}$

The force will be maximum $F_{\max }=q \vee B$

 

(e) Work done by force due to magnetic field in motion of a charged particle is always zero.

When a charged particle move in a magnetic field, then force acts on it is always perpendicular to displacement,

so the work done, $\left.\quad \mathrm{W}=\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{ds}}=\int \mathrm{F} d \mathrm{s} \cos 90^{\circ}=0 \quad \text { (as } \theta=90^{\circ}\right)$

And as by work-energy theorem $\mathrm{W}=\Delta \mathrm{KE},$ the kinetic energy $\left(=\frac{1}{2} \mathrm{mv}^{2}\right)$, remains unchanged and hence speed of charged particle v remains constant.

However, in this situation the force changes the direction of motion, so the direction of velocity of $\vec{v}$ the charged particle changes continuously.

 

(f) For motion of charged particle in a magnetic field $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$

 

So magnetic induction $\overrightarrow{\mathrm{B}}$ can be defined as a vector having the direction in which a moving charged particle does not experience any force in the field and magnitude equal to the ratio of the magnitude of maximum force to the product of magnitude of charge with velocity

Difference in Force on a Charged Particle by Magnetic Field and Electric Field

 

 

Also Read:

Biot Savart’s Law

 

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Motion of Charged Particle in a Magnetic Field | Moving Charges and Magnetism Class 12, JEE & NEET

When a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences a force that reduces the component of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a Magnetic Mirror. Motion of a charged particle in magnetic field is characterized by the change in the direction of motion. It is expected also as magnetic field is capable of only changing direction of motion. In order to keep the context of study simplified, we assume magnetic field to be uniform. This assumption greatly simplifies the description and lets us easily visualize the motion of a charged particle in magnetic field.

Motion of a Charged Particle in a Magnetic Field

Motion of a charged particle when it is moving collinear with the field magnetic field is not affected by the field (i.e. if motion is just along or opposite to magnetic field) ( : $: \quad F=0$ ) Only the following two cases are possible:

 

Case I: When the charged particle is moving perpendicular to the field The angle between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{v}}$ is $\theta=90^{\circ}$

So the force will be maximum ( $=$ qvB ) and always perpendicular to motion (and also field);

Hence the charged particle will move along a circular path (with its plane perpendicular to the field).

Centripetal force is provided by the force qvB,

n case of circular motion of a charged particle in a steady magnetic field :

i.e., with increase in speed or kinetic energy, the radius of the orbit increases. For uniform circular motion $v=\omega r$

Angular frequency of circular motion, called cyclotron or gyro-frequency. $\omega=\frac{\mathrm{v}}{\mathrm{r}}=\frac{\mathrm{qB}}{\mathrm{m}}$

and the time period, $\quad \mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \frac{\mathrm{m}}{\mathrm{qB}}$

i.e., time period (or frequency) is independent of speed of particle and radius of the orbit.

Time period depends only on the field B and the nature of the particle,

i.e., specific charge (q/m) of the particle.

This principle has been used in a large number of devices such as cyclotron (a particle accelerator), bubble-chamber (a particle detector) or mass-spectrometer etc.

motion of charged particle in electric and magnetic field

 

case II : The charged particle is moving at an angle $\theta$ to the field :

$\left(\theta \neq 0^{\circ}, 90^{\circ} \text { or } 180^{\circ}\right)$

Resolving the velocity of the particle along and perpendicular to the field.

The particle moves with constant velocity v cos $\theta$ along the field ($\because$ no force acts on a charged particle when it moves parallel to the field).

And at the same time it is also moving with velocity $v$ sin $\theta$ perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field)

So the resultant path will be a helix with its axis parallel to the field $\overrightarrow{\mathrm{B}}$ as shown in fig. The pitch p of the helix $=$ linear distance travelled in one rotation

$p=T(v \cos \theta)=\frac{2 \pi m}{q B}(v \cos \theta)$

 

Also Read:

Biot Savart’s Law

 

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Torque on a Current loop in a Magnetic Field || Moving Charges and Magnetism – Class 12, JEE & NEET

Without torque, there would be no twists and turns, no spins! Wouldn’t life be boring that way? Torque gives a rotational motion to an object that would otherwise not be possible. So here we will study about Torque and Torque on a Current loop in a Magnetic Field in detailed derivation.

Torque Experienced by a Current Loop in a Uniform Magnetic Field

Let us consider a rectangular current loop PQRS of sides a and b suspended vertically in a uniform magnetic field . $\overrightarrow{\mathrm{B}}$. Let $\theta$ be the angle between the direction of $\overrightarrow{\mathrm{B}}$ and the vector perpendicular to the plane of the loop.

Let us first consider the two straight parts PQ and RS.

The forces acting on these parts are clearly equal in magnitude and opposite in direction. These forces are collinear, so, the net force or net torque due to this pair of forces is zero.

The forces on the straight current segments SP and QR are

(i) equal in magnitude (ii) opposite in direction (iii) not collinear.

These forces shown in figure (b).

This pair of forces produces a torque whose lever arm is b sin\theta. Magnitude of each force $=\mathrm{B} \mathrm{I}$ a

The magnitude of the torque $\vec{\tau}$ is given by

where $A$ is the area of the coil.

But $\mathrm{IA}=\mathrm{M}$ (magnitude of magnetic dipole moment)

The direction of the torque is vertically downwards along the axis of suspension [dotted line in fig (b)]

It will rotate the loop clockwise about its axis.

Note 1. If the rectangular loop has N turns, then the torque increases N times and becomes $\mathrm{N} \mathrm{B}$ I $\mathrm{A} \sin \theta$

 

Note 2

$\mathrm{eq}^{\mathrm{n}}$ (i) could also be written as $\vec{\tau}=\mathrm{I}(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})$ OR $\vec{\tau}=\mathrm{I} \mathrm{A} \hat{\mathrm{n}} \times \overrightarrow{\mathrm{B}}$

where $\hat{\mathrm{n}}$ is a unit vector normal to the plane of the loop.

 

Also Read:

Biot Savart’s Law

 

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Flemings Left Hand Rule Class 12 Physics

Flemings Left Hand Rule Class 12 states that if we stretch the thumb, the forefinger and the middle finger of our left hand such that they are mutually perpendicular to each other.
If the forefinger gives the direction of current and middle finger points in the direction of magnetic field then the thumb points towards the direction of the force or motion of the conductor.

Then if the fore-finger points in the direction of field $(\overrightarrow{\mathrm{B}})$, the central finger in the direction of current , the thumb will point in the direction of force.

(ii) Right-hand Palm Rule : Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point in the direction of field $\overrightarrow{\mathrm{B}}$ and thumb in the direction of current $\mathbf{I}$, the normal to palm will point in the direction of force.

Regarding the force on a current-carrying conductor in a magnetic field it is worth mentioning that :

(a) As the force BI d\ell sin \theta is not a function of position r, the magnetic force on a current element is non-central [a central force is of the form $\mathrm{F}=\mathrm{Kf}(\mathrm{r}) \overrightarrow{\mathrm{n}_{\mathrm{r}}}]$

(b) The force d\vec $\overrightarrow{\mathrm{F}}$ is always perpendicular to both $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ though $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ may or may not be perpendicular to each other.

(c) In case of current-carrying conductor in a magnetic field if the field is uniform i.e.,

$\overrightarrow{\mathrm{B}}=$ constt.,

$\overrightarrow{\mathrm{F}}=\int \mathrm{I} \mathrm{d} \vec{\ell} \times \overrightarrow{\mathrm{B}}=\mathrm{I}\left[\int \mathrm{d} \vec{\ell}\right] \times \overrightarrow{\mathrm{B}}$

and as for a conductor $\int_{\mathrm{d}} \vec{\ell}$ represents the vector sum of all the length elements from initial to final point, which in accordance with the law of vector addition is equal to the length vector $\overrightarrow{\ell^{\prime}}$ joining initial to final point, so a current-carrying conductor of any arbitrary shape in a uniform field experience a force

$\vec{F}=I\left[\int d \vec{\ell}\right] \times \vec{B}=I \ell^{\prime} \times \vec{B}$

where $\vec{\ell}$ is the length vector joining initial and final points of the conductor as shown in fig.

(d) If the current-carrying conductor in the form of a loop of any arbitrary shape is placed in a uniform field,

$\overrightarrow{\mathrm{F}}=\oint \mathrm{Id\vec{\ell }} \times \overrightarrow{\mathrm{B}}=\mathrm{I}[\oint \overrightarrow{\mathrm{d} \vec{\ell}}] \times \overrightarrow{\mathrm{B}}$

and as for a closed loop, $\oint \mathrm{d} \vec{\ell} \text { is always zero. [vector sum of all } \mathrm{d} \vec{\ell}]$

i.e., the net magnetic force on a current loop in a uniform magnetic field is always zero as shown in fig.

Here it must be kept in mind that in this situation different parts of the loop may experience elemental force due to which the loop may be under tension or may experience a torque as shown in fig.

Current Loop in a Uniform Field

(e) if a current-carrying conductor is situated in a non-uniform field, its different elements will experience different forces; so in this situation,

$\overrightarrow{\mathrm{F}_{\mathrm{R}}} \neq 0$ but $\overrightarrow{\mathrm{\tau}}$ may or may not be zero


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Force on a Current Carrying Conductor in a Magnetic Field – Class 12, JEE & NEET

Force on a Current Carrying Conductor in Magnetic field: A current carrying conductor produces a magnetic field around it. i.e. behaves like a magnet and exerts a force when a magnet is placed in its magnetic field. Similarly a magnet also exerts equal and opposite force on the current carrying conductor. The direction of this force can be determined using Flemings left-hand rule.

Current Element- – Current element is defined as a vector having magnitude equal to the product of current with a small part of length of the conductor and the direction in which the current

is flowing in that part of the conductor.

With the help of experiments Ampere established that when a current element 1 de is placed

in a magnetic field $\overrightarrow{\mathrm{B}},$ it experiences a force

$\overrightarrow{\mathrm{dF}}=\mathrm{I} \overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$

The magnitude of force is $\quad \mathrm{dF}=\mathrm{B} \mathrm{I}$ d\ell $\sin \theta$

$[\theta \text { is the angle between the } \overrightarrow{\mathrm{d} \ell} \text { and } \overrightarrow{\mathrm{B}}]$

(i) When $\sin \theta=\min =0,$ i.e., $\theta=0^{\circ}$ or $180^{\circ}$

Then force on a current element $=0$ (minimum)

A current element in a magnetic field does not experience any force if the current in it is collinear with the field.

(ii) When $\sin \theta=\max =1,$ i.e., $\theta=90^{\circ}$

The force on the current will be maximum ( $=$ BI d\ell)

Force on a current element in a magnetic field is maximum $(=\text { BI } \mathrm{d} \ell)$

When it is perpendicular to the field.

The direction of force is always perpendicular to the plane containing I $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{B}}$.

The direction of force on current element I $\mathrm{d} \vec{\ell}$ and $\overrightarrow{\mathrm{B}}$ are perpendicular to each other can also be determined by applying either of the following rules:

1. Fleming’s Left Hand Rule

2. Right Hand Palm Rule

 

 

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Solenoid and Toroid | Difference between Solenoid and Toroid

A toroid works as an inductor, which boosts the frequency to appropriate levels. Inductors are electronic components that are passive, so that they can store energy in the form of magnetic fields. A toroid turns, and with those turns induces a higher frequency. Toroids are more economical and efficient than solenoids. So here we will study in detail about Solenoid and Toroids and the differences between both.

  1. Solenoid
  2. Toroids
  3. Difference between Solenoid and Toroid
  4. Similarities Between Solenoid and Toroid

Toroid :

A toroid is an endless solenoid in the form of a ring, as shown in fig.

A toroid is often used to create an almost uniform magnetic field in some enclosed area.

The device consists of a conducting wire wrapped around a ring (a torus) made of a non-condunting material.

Let a toroid having N closely spaced turns of wire,

magnetic field in the region occupied by the torus = B

Radius of the Toroid ring = R

To calculate the field, we must evoluate $\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}$ over the circle of radius $\mathrm{R} .$ By symmetry magnitude of the field is constant at the circumference of the circle and tangent to it.

So, $\quad \quad \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} \vec{\ell}=\mathrm{B} \ell=\mathrm{B}(2 \pi \mathrm{R})$

This result shows that $\mathrm{B} \propto \frac{1}{\mathrm{R}}$ and hence is nonuniform in the region occupied by torus. However,

if R is very large compared with the cross-sectional radius of the torus, then the field is approximately uniform inside the torus.

Number of turns per unit length of torus $n=\frac{N}{2 \pi R}$

$\therefore \quad B=\mu_{0} n I$

For an ideal toroid, in which turns are closely spaced, the external magnetic field is zero. This is because the net current passing through any circular path laying outside the torid is zero. Therefore, from Ampere’s law we find that $B=0,$ in the regions exterior to the torus.

Solenoid:

A long wire wrapped around in helical shape is known as Solenoid. They are cylindrical in shape as you can see in the image above. A solenoid is used in experiments and research field regarding the magnetic field.
The magnetic field inside solenoid is uniform.

Difference between Solenoid and Toroid

Solenoid and toroid both work on the principle of electromagnetism and both behave like an electromagnet when current is passed. The magnetic field produces by them is $\mathrm{B}=$ Uo.ni. Even after having so many similarities solenoid and toroid differs in property such as shape:

Solenoid

Toroid

Cylindrical in shape Circular in shape
Magnetic field is created outside Magnetic field is created within
Magnetic field is created outside It does not have a uniform magnetic field inside it.
Has uniform magnetic field inside it. Magnetic feild Outside :- magnetic field (B) $=0$
Magnetic feild due to solenoid is $(B)=$ Uo.nì Magnetic feild Inside:-Magnetic field $(B)=0$
Magnetic feild Within the toroid:- Magnetic field (B) =
Uo.ni

 

Similarities between Solenoid and Toroid:

1. Both works on the principle of Electromagnetism.

2. When current is passed through them, they both act like an electromagnet.

3. Magnetic field due to the solenoid and within the toroid is the same. B = Uo.ni

 

 

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Magnetic Field of a Solenoid || Class 12 Physics Notes

The Solenoid is a coil of wire that acts like an electromagnet when a flow of electricity passes through it. Electromagnetic solenoids find uses all over the world. You can hardly swing a bat without hitting a solenoid.Speakers and microphones both contain solenoids. In fact, a speaker and microphone are pretty much exactly the same things in reverse of each other. So here we will study in detail about Magnetic Field of a Solenoid Derivation in this article:

Magnetic field of a Solenoid

A solenoid is a long cylindrical helix. It is made by winding closely a large number of turns of insulated copper wire over a tube of card-board or china-clay. When electric current is passed through the solenoid, a magnetic field is produced around and within the solenoid.

Figure shows the lines of force of the magnetic field due to a solenoid. The lines of force inside the solenoid are nearly parallel which indicate that the magnetic field ‘within’ the solenoid is uniform and parallel to the axis of the solenoid.

Magnetic field of a solenoid derivation

Let there be a long solenoid of radius ‘a’ and carrying a current I. Let n be the number of turns per unit length of the solenoid. Let $P$ be a point on the axis of the solenoid.

Let us imagine the solenoid to be divided up into a number of narrow coils and consider one such coil AB of width $\delta x$. The number of turns in this coil is n\deltax. Let $x$ be the distance of the point $P$ from the centre $\mathrm{C}$ of this coil. The magnetic field at $\mathrm{P}$ due to this elementary coil is given by

Let average distance of circumference is r and $\delta \theta$ the angle subtended by the coil at P.

$$ \sin \theta=\frac{\mathrm{BN}}{\mathrm{AB}}=\frac{\mathrm{r} \delta \theta}{\delta \mathrm{x}} \quad \text { or } \quad \delta \mathrm{x}=\frac{\mathrm{r} \delta \theta}{\sin \theta} $$

In $\Delta \mathrm{ACP},$ we have $\quad \mathrm{a}^{2}+\mathrm{x}^{2}=\mathrm{r}^{2}$

$$ r^{3}=\left(a^{2}+x^{2}\right)^{3 / 2} $$

Substiuting these values of $\delta x$ and $\left(a^{2}+x^{2}\right)^{3 / 2}$ in eq. (i), we get

$$ \delta \mathrm{B}=\frac{\mu_{0}(\mathrm{nr} \delta \theta) \mathrm{I} \mathrm{a}^{2}}{2 \mathrm{r}^{3} \sin \theta}=\frac{\mu_{0} \mathrm{nI} \mathrm{a}^{2}}{2 \mathrm{r}^{2}} \frac{\delta \theta}{\sin \theta} $$

The magnetic field $\mathrm{B}$ at $\mathrm{P}$ due to the whole solenoid can be obtained by integrating the above expression between the limits $\theta_{1}$ and $\theta_{2},$ where $\theta_{1}$ and $\theta_{2}$ are the semi-vertical angles subtended at $P$ by the first and the last turn of the solenoid respectively. Thus

or $\quad \mathrm{B}=\frac{1}{2} \mu_{0} \mathrm{n} \mathrm{I}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ ……….(ii)

If the observation point P is well inside a very long solenoid

Thus, the magnetic field at the ends of a ‘long’ solenoid is half of that at the center. If the solenoid is sufficiently long, the field within it (except near the ends) in uniform. It does not depend upon the length and area of cross-section of the solenoid. Just as a parallel-plate capacitor produces uniform and known electric field, a solenoid produces a uniform and known magnetic field.

The ‘uniform’ magnetic field within a long solenoid is parallel to the solenoid axis.

Its direction along the axis is given by a curled-straight right-hand rule. “If we grasp the solenoid with our right hand so that our fingers follow the direction of the current in the winding’s, then out extended right thumb will point in the direction of the axial magnetic field”.

 

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Ampere’s Circuital Law and it Applications || Magnetic Effects of Current Class 12

Do you know about the Ampere’s Circuital Law, well it is a current distribution which helps us to calculate the magnetic field. And yes, the Biot-Savart’s law does the same but Ampere’s law uses the case high symmetry. We will first understand the ampere’s circuital law, its definition, formulae, & Applications of Ampere’s Law in detail,

  1. What is Ampere’s Circuital Law?
  2. Application of Ampere’s Circuital Law

Ampere’s Circuital Law

Ampere’s circuital law states that the line integral of magnetic field induction $\overrightarrow{\mathrm{B}}$ around any closed path in vacuum is equal to $\mu_{0}$ times the total current threading the closed path, i.e.,

This result is independent of the size and shape of the closed curve enclosing a current.

This is known as Ampere’s circuital law.

Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.

Ampere’s law may be derived from the Biot-Savart law and Biot-Savart law may be derived from the Ampere’s law.

Ampere’s law is more useful under certain symmetrical conditions.

Biot-Savart law based on the experimental results whereas Ampere’s law based on mathematical.

Applications of Ampere’s Law

(a) Magnetic induction due to a long current carrying wire.

Consider a long straight conductor Z-Z’ is along z-axis. Let I be the current flowing in the direction as shown in Fig. The magnetic field is produced around the conductor. The magnetic lines of force are concentric circles in the XY plane as shown by dotted lines. Let the magnitude of the magnetic field induction produced at a point P at distance r from the conductor is

Consider a close circular loop as shown in figure.

According to Ampere’s law $\oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d}} \ell=\mu_{0} \sum \mathrm{I}$

The direction of $\overrightarrow{\mathrm{B}}$ at every point is along the tangent to the circle.

Consider a small element $\overrightarrow{\mathrm{d} \ell}$ of the circle of radius r at P. The direction of $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{d} \ell}$ the same. Therefore, angle between them is zero.

Line integral of $\overrightarrow{\mathrm{B}}$ around the complete circular path of radius $\mathrm{r}$ is given by

$\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} \ell=\oint \mathrm{B} \mathrm{d} \ell \cos 0^{\circ}$ $=\quad \mathrm{B} \oint \mathrm{d} \ell=\mathrm{B} \times 2 \pi \mathrm{r}$ $(\oint \mathrm{d} \ell=2 \pi \mathrm{r}=$ cicumference of the circle.) and $\quad \sum I=I$

So we get $\mathrm{B} \times 2 \pi \mathrm{r}=\mu_{0} \mathrm{I}$

(b) Magnetic field created by a long current carrying conducting cylinder

A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross-section of the wire.

For finding the behavior of magnetic field due to this wire, let us divide the whole region into two parts.

(a) $\mathrm{r} \geq \mathrm{R}$ and

(b) $\mathrm{r}<\mathrm{R}$

$r=$ distance from the centre of the wire.

For $\mathrm{r} \geq \mathrm{R}:$ For closed circular path denoted by ( 1) from symmetry $\overrightarrow{\mathrm{B}}$ must be constant in magnitude and parallel to $\overrightarrow{\mathrm{d} \ell}$ at every point on this circle. Because the total current passing through the plane of the circle is I.

For $\mathrm{r}<\mathrm{R}:$ The current $\mathrm{I}$ passing through the plane of circle 2 is less than the total current I. Because the current is uniform over the cross-section of the wire.

Current through unit area $=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}}$

So current through area enclosed by circle 2 is $\mathrm{I}^{\prime}=\frac{\mathrm{I} \pi \mathrm{r}^{2}}{\pi \mathrm{R}^{2}}$

Now we apply Ampere’s law for circle $2 .$

The magnitude of the magnetic field versus $r$ for this configuration is plotted in figure. Note that inside the wire $\mathrm{B} \rightarrow 0$ as $\mathrm{r} \rightarrow 0 .$ Note also that eqn. (a) and eqn (b) give the same value of the magnetic field at $r=R,$ demonstrating that the magnetic field is continuous at the surface of the wire.

(c) Magnetic field due to a conducting current carrying hollow cylinder

Consider a conducting hollow cylinder with inner radius $r_{1}$ and outer radius $r_{2} .$ And current $\mathrm{I}$ is flowing through it.

(I) $\quad$ For $r<r_{1}$

$\sum \mathrm{I}=0$ and hence $\quad B=0$

(II) $\quad$ For $r_{1}<r<r_{2}$

Now current I is flowing through area $\left[\pi r_{2}^{2}-\pi r_{1}^{2}\right]$

So, current per unit area $=\frac{I}{\pi\left(r_{2}^{2}-r_{1}^{2}\right)}$

$\therefore$ current flowing through area in bet” $\mathrm{r}_{1}<\mathrm{r}<\mathrm{r}_{2}$ is $\mathrm{I}=\frac{\mathrm{I}}{\pi\left(\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}\right)} \times\left(\pi \mathrm{r}^{2}-\pi \mathrm{r}_{1}^{2}\right)$

by using ampere’s law for circle of radius $\mathrm{r} \oint \overrightarrow{\mathrm{B}} . \overrightarrow{\mathrm{d}} \vec{\ell}=\mu_{0} \sum \mathrm{I}$

or $\quad \oint B d \ell \cos 0^{\circ}=\mu_{0}\left[\frac{I\left(r^{2}-r_{1}^{2}\right)}{r_{2}^{2}-r_{1}^{2}}\right]$

or $\quad \mathrm{B} \oint \mathrm{d} \ell=\mu_{0} \mathrm{I}\left[\frac{\mathrm{r}^{2}-\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}}\right]$

or $\quad B=\frac{\mu_{0} I}{2 \pi r}\left[\frac{r^{2}-r_{1}^{2}}{r_{1}^{2}-r_{1}^{2}}\right]$

$[\because \oint \mathrm{d} \ell=2 \pi \mathrm{r}]$

(a) For $r=r_{2}$

$B=\frac{\mu_{0} I}{2 \pi r_{2}}$

(b) For $r>r_{2}$

$B=\frac{\mu_{0} I}{2 \pi r}$

 

 

 

Also Read:

Biot Savart’s Law

 

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Magnetic Field at the Centre of a Circular Coil | Circular Current loop as a Magnetic Dipole

Here we will study about the Magnetic field at the center of a circular coil. Here we will study about the number of cases

Magnetic Field at the Center of a Circular Current-Carrying Coil

Consider a circular coil of radius r through which current I is flowing. Let AB be an infinitesimally small element of length d\ell. According to Biot-Savart’s law, the magnetic field dB at the centre P of the loop due to this small element d $\ell$ is

                                                           

$\mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{r^{2}}$ where $\theta$ is the angle between $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$.

$\left.\therefore \quad \mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d \ell \sin 90^{\circ}}{r^{2}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d \ell}{\mathrm{r}^{2}} \quad \text { (for circular loop, } \theta=90^{\circ}\right)$

The loop can be supposed to consists of a number of small elements placed side by side. The magnetic field due to all the elements will be in the same direction. So, the net magnetic field at P is given by

                                                           

$\mathrm{B}=\sum \mathrm{dB}=\sum \frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} \mathrm{d} \ell}{\mathrm{r}^{2}}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}^{2}} \sum \mathrm{d} \ell$

$\therefore \quad \mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi r^{2}} \times 2 \pi r$

                                           

$| \Sigma \mathrm{d} \ell=\text { circumference of the circle }=2 \pi \mathrm{r})$

       

 

 

Magnetic Field due to part of current carrying circular conductor (Arc) :


$B=\frac{\mu_{0} I}{4 \pi r^{2}} \sum d \ell\left(\because \frac{\sum d \ell}{r}=\alpha\right)$

$B=\frac{\mu_{0} I}{4 \pi r} \alpha$

 

Magnetic Field on the Axis of a Circular Coil

– Consider a circular loop of radius a through which current I is flowing as shown in fig. The point $P$ lies on the axis of the circular current loop i.e., along the line perpendicular to the plane of the loop and passing through its centre.

Let $x$ be the distance of the observation point $P$ from the centre 0 of the loop. Let us consider an infinitesimally small element $\mathrm{AB}$ of length d\ell. Radius of the loop $=\mathrm{a}$ According to Biot-Savart’s law, the magnetic field at P due to this small element

$\begin{aligned} \overrightarrow{\mathrm{dB}} &=\frac{\mu_{0} \mathrm{I}}{4 \pi r^{3}}[\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{r}}] \\ \mathrm{dB} &=\frac{\mu_{0} \mathrm{I} \mathrm{d} \ell \sin \theta}{4 \pi \mathrm{r}^{2}} \end{aligned}$

or $\quad \mathrm{dB}=\frac{\mu_{0} \mathrm{I} \mathrm{d} \ell}{4 \pi \mathrm{r}^{2}}\left(\theta=90^{\circ}\right)$

The direction of $\overrightarrow{\mathrm{dB}}$ is perpendicular to the plane of the current element $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}(\mathrm{CP})$ as shown in fig. by $\overrightarrow{\mathrm{PM}}$

Similarly if we consider another small element just diametrically opposite to this element then

magnetic field due to this at point $P$ is $\overrightarrow{\mathrm{dB}^{\prime}},$ denoted by PN and of the same magnitude. $\mathrm{d} \mathrm{B}=\mathrm{dB}^{\prime}$

Both $\overrightarrow{\mathrm{dB}}$ and $\overrightarrow{\mathrm{dB}^{\prime}}$ can be resolved into two mutually perpendicular components along $\mathrm{PX}$ and $\mathrm{zz}$ :

The components along $z z^{\prime}\left[\mathrm{d} \mathrm{B} \cos \alpha \text { and } \mathrm{dB}^{\prime} \text { cos\alpha }\right]$ cancel each other as they are equal and opposite in direction.

The same will hold for such other pairs of current elements. over the entire circumference of the loop.

Therefore, due to the various current elements, the components of magnetic field is only along PX will contribute to the magnetic field due to the whole loop at point P.

 

Magnetic dipole moment of the current loop

The current loop can be regarded as a magnetic dipole which produces its magnetic field and magnetic dipole moment of the current loop is equal to the product of ampere turns and area of current loop. we can write

Case II : If the observation point is far far away from the coil, then $a<<x$. So, $a^{2}$ can be neglected in comparison to $x^{2}$.

$\therefore \quad B=\frac{\mu_{0} N I a^{2}}{2 x^{3}}$

terms of magnetic dipole moment, $\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{x}^{3}} \quad\left[\mathrm{B}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{NIA}}{\mathrm{x}^{3}}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{NIA}}{\mathrm{x}^{3}}\right]$

 

Also Read:

Biot Savart’s Law

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

 

Helmholtz Coils | Magnetic Field between two Coils

Helmholtz coil is named after the German physicist Hermann von Helmholtz. It is comprised of two identical magnetic coils positioned in parallel to each other, and their centers are aligned in the same x-axis. The two coils are separated by a distance equal to the radius like a mirror image as shown in Figure 1. When current is passing through the two coils in the same direction, it generates a uniform magnetic field in a three-dimension region of space within the coils. Helmholtz coils are normally used for scientific experiments, magnetic calibration, to cancel background (earth’s) magnetic field, and for electronic equipment magnetic field susceptibility testing.

Helmholtz Coils

The two coaxial coils of equal radii placed at distance equal to the radius of either of the coils and in which same current in same direction is flowing are known as Helmholtz coils.

For these coils $x=\frac{a}{2}, I_{1}=I_{2}=I, a_{1}=a_{2}=a$

The two coils are placed mutually parallel to each other these coils are used to produce a uniform magnetic field. In between two coils along the axis at middle point rate of change of magnetic field is constant so if distance increases from a coil magnetic field decreases but distance from the another coil decreases so magnetic field due to second coil increases and hence the resultant magnetic field produced in the region between two coils remains uniform.

or $\mathrm{B}=0.76 \frac{\mu_{0} n \mathrm{I}}{a} \quad$ or $\mathrm{B}=1.423 \mathrm{B}_{\mathrm{C}}$

(Bc is magnetic field at the centre of a single coil.)\

 

 

Also Read:

Biot Savart’s Law

 

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Right Hand Palm Rule

So far we have described the magnitude of the magnetic force on a moving electric charge, but not the direction. The magnetic field is
a vector field, thus the force applied will be oriented in a particular direction. There is a clever way to determine this direction using nothing more than your right hand. The direction of the magnetic force $F$ is perpendicular to the plane formed by $v$ and $B$, as determined by the right hand palm rule, which is illustrated in the figure. The right hand rule states that ito determine the direction of the magnetic force on a positive moving charge, $f$, point the thumb of the right hand in the direction of $v,$ the fingers in the direction of $B,$ and a perpendicular to the palm points in the direction of $F$

Right Hand Palm Rule

If we hold the thumb of right hand mutually perpendicular to the grip of the fingers such that the curvature of the finger represents the direction of current in the wire loop, then the thumb of the right hand will point in the direction of magnetic field near the centre of the current loop

Graph of B v/s x

As soon as x increases magnetic field B decreases, dependence of B on x is shown in figure

Rate of change of B with respect to x is different at different values of x

for $x<\pm \frac{a}{2}$ curve is convex and $\quad$ for $x>\pm \frac{a}{2}$ curve is concave

At $x=\pm \frac{a}{2} \quad$ we get $\frac{d B}{d x}=$ const $,$ and $\frac{d^{2} B}{d x^{2}}=0$

So at $x=+\frac{a}{2} \&-\frac{a}{2}$ B varies linearly with $x$

These points are called point of inflexion.

Distance in between these two points is equal to radius of the coil

$B=\frac{B_{C}}{\left(1+\frac{x^{2}}{a^{2}}\right)^{3 / 2}}$

$\because$ Magnetic field at the centre of coil $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{NI}}{2 \mathrm{a}}$

 

 

Also Read:

Biot Savart’s Law

 

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Magnetic Field due to Infinite Straight Conductor | Magnetic Effects of Current Class 12

Magnetic field due to long straight conductor

Here we will discuss all the cases involved in magnetic field due to Conductor such as Magnetic Field due to Infinite Straight Conductor, and many more discussed below:

Consider a long straight conductor XY through which current I is flowing from $X$ to Y. Let $P$ be the observation point at a distance ‘r’ ‘from the conductor XY. Let us consider an infinitesimally small current element $\mathrm{CD}$ of length d\ell. Let s be the distance of $\mathrm{P}$ from the mid-point $\odot$ of the current element. Let $\theta$ be the angle which OP makes with the direction of current. The magnetic field at $P$ due to the current element $C D$ is $\mathrm{dB}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{\mathrm{s}^{2}}[\text { Biot-Savart’s law }]$

The magnetic field at P due to the whole of the conductor XY

Case I :

If the conductor is infinitely long, then $\theta_{1}=90^{\circ}$ and $\theta_{2}=90^{\circ}$

$\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi}\left[\sin \frac{\pi}{2}+\sin \left(\frac{\pi}{2}\right)\right]=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}[1+1]=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}$

Or

Case II : If conductor is of infinite length but one end is in front of point $P$ i.e. one end of conductor starts from point $N$ then $\theta_{1}=0^{\circ}$ and $\theta_{2}=90^{\circ}$

                                             

Case III :

Conductor is finite length and point P is just in front of middle of the conductor

Case IV :

 

 

Also Read:

Biot Savart’s Law

 

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Right hand Thumb Rule in Physics Class 12 | Magnetic Effects of Current

The right hand thumb rule in Physics states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

Right Hand Thumb Rule:

If we grasp the conductor in the palm of the right hand so that the thumb points in the direction of the flow of current, then the direction in which the fingers curl, gives the direction of magnetic field lines.

For the current flowing through the conductor in the direction shown in fig. (a) or (b), both the rules predict that magnetic field lines will be in anticlockwise direction, when seen from above.

The magnetic field produced by a current-carrying straight conductor is of circular symmetry. The magnetic lines of force are concentric circles with the current carrying conductor passing through their common center. The plane of the magnetic lines of force is perpendicular to the length of the conductor.

 

Also Read:

Biot Savart’s Law

 

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Magnetic Effect of Electric Current Class 12 Notes | Introduction

The electricity and magnetism are linked to each other and it is proved when the electric current passes through the copper wire, it produces a magnetic effect. The electromagnetic effects first time noticed by Hans Christian Oersted. Oersted discovered a magnetic field around a conductor carrying electric current. Magnetic field is a quantity, which has both magnitude and direction. The direction of a magnetic field is usually taken to be the direction in which, a north pole of the compass needle moves inside it. So here you will get Magnetic Effect of Electric Current Class 12 complete Notes to prepare for Boards as well as for JEE & NEET Exams.

  1. Introduction
  2. Unit of Magnetic Field
  3. Biot Savart’s Law
  4. Biot Savart’s law in Vector Form

 

Oersted discovered a magnetic field around a conductor carrying electric current. Other related facts are as follows:

(a) A magnet at rest produces a magnetic field around it while an electric charge at rest produce an electric field around it.

(b) A current carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving with a uniform velocity has an electric as well as a magnetic field around it.

(c) An electric field cannot be produced without a charge whereas a magnetic field can be produced without a magnet.

(d) No poles are produced in a coil carrying current but such a coil shows north and south polarities.

(e) All oscillating or an accelerated charge produces E.M. waves also in additions to electric and magnetic fields.

                                                 

                                                 

 

Unit of Magnetic field

UNIT OF $\overrightarrow{\mathrm{B}}:$ MKS weber/metre $^{2},$ SI tesla, CGS maxwell cm’ or gauss.

One Tesla $=$ one (weber/m’) $=10^{4}$ (maxwell/cm’) $=10^{4}$ gauss

Biot-Savart’s Law

With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field $\overrightarrow{\mathrm{d} B}$ at a point $P$ associated with a length element $\overrightarrow{\mathrm{d} \ell}$ of a wire carrying a steady current I. 

            

$\mu_{0}$ is called permeability of free space $\frac{\mu_{0}}{4 \pi}=10^{-7}$ henry/meter.

$$ 1(\mathrm{H} / \mathrm{m})=1 \frac{\mathrm{Tm}}{\mathrm{A}}=1 \frac{\mathrm{Wb}}{\mathrm{Am}}=1 \frac{\mathrm{N}}{\mathrm{A}^{2}}=1 \frac{\mathrm{Ns}^{2}}{\mathrm{c}^{2}} $$

DIMENSIONS of $\mu_{0}=\left[\mathrm{M}^{\prime} \mathrm{L}^{\prime} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$

For vaccum $: \sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}=\mathrm{c}=3 \times 10^{8} \mathrm{m} / \mathrm{s}$

Biot-Savart law in vector form

                     

[Note: Static charge is a source of electric field but not of magnetic field, whereas the moving charge is a source of electric field as well as magnetic field.]

the direction of $\mathrm{d} \mathrm{B}$ is perpendicular to the plane determined by $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ (i.e. if $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{r}}$ lie in the plane of the paper then $\overrightarrow{\mathrm{dB}}$ is $\perp$ to plane of the paper). In the figure, direction of

$\overrightarrow{\mathrm{dB}}$ is into the page. (Use right hand screw rule).

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Difference Between Potentiometer and Voltmeter – Current Electricity || Class 12, JEE & NEET

The Potentiometer and the Voltmeter both are the voltage measuring devices. The main Difference Between Potentiometer and Voltmeteris that the potentiometer measures the emf of the circuit whereas voltmeter measures the end terminal voltage of the circuit. The other differences between the potentiometer and the voltmeter are explained below in the comparison chart.

  1. Stabdardization of Potentiometer
  2. Key Differences

Comparison Chart

Difference between Potentiometer and Ammeter

 

Standardization of Potentiometer

The process of determination of potential gradient on wire of potentiometer is known as standardisation of potentiometer. A standard cell is one whose emf remains constant. Cadmium cell with emf 1.0186 V at $20^{\circ}$C is used as a standard cell. In laboratory a Daniel cell with emf 1.08 V is usually used as a standard cell.

standardization of Patentiometer
If $\ell_{0}$ is the balancing length for standard emf $E_{0}$ then potential gradient $x=\frac{E_{0}}{l_{0}}$

Standardization of Potentiometer

Key Differences:

The following are the key differences between Potentiometer and voltmeter.

  1. The Potentiometer is an instrument used for measuring the emf, whereas the voltmeter is a type of meter which measures the terminal voltage of the circuit.
  2. The Potentiometer accurately measures the potential difference because of zero internal resistance. Whereas, the voltmeter has a high internal resistance which causes the error in measurement. Thus the voltmeter approximately measures the voltage.
  3. The sensitivity of the Potentiometer is very high, i.e. it can measure small potential differences between the two points. The voltmeter has low sensitivity.
  4. The Potentiometer uses the null deflection type instrument whereas the voltmeter uses the deflection type instrument.
  5. The Potentiometer has infinite internal resistance, whereas the Potentiometer has high measurable resistance.

Conclusion

The Potentiometer and voltmeter both measures the emf in volts. The Potentiometer is used in a circuit where the accurate value of voltage is required. For approximate calculation, the voltmeter is used.

 

 

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

 

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

 

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  • a valid e-mail id
  • valid and active mobile number
  • past academic mark sheets and certificates

The candidates have to upload Photo and Signatue in the application form of JEE Main 2020 and as per the given specifications:

Scanned image Format Size Remarks
Photo JPG/JPEG 10 kb–200 kb Colored or black/white
(but clear contrast)
Signature JPG/JPEG 4 kb–30 kb Must be in Running hand

 

JEE Main April 2020 Important Points

Registration Process will be different for those who applied in January and for those haven’t applied in January JEE Main Exam.
If you have applied in JEE Main January 2020: No need to register. You can directly login with the same credentials and fill the application form for JEE Main April attempt.

If you haven’t applied in January 2020 JEE Main Exam: Then you have to fill the application form. First you need to register and then fill the correct details in the form, upload images in the application form & pay the application fee.

JEE Main April 2020 Exam Pattern

As per the eligibility criteria for, a few changes in the pattern of the question paper(s) and number of question(s) for B.E./B.Tech has been approved by the JEE Apex Board (JAB) for the conduct of JEE (Main)-2020 Examination.

 

Paper Subject with Number of Questions Type of Questions TIMING OF THE EXAMINATION(IST)
B.tech Mathematics – 25(20+5)

Physics – 25(20+5)

Chemistry – 25(20+5)

20 Objective Type – Multiple Choice Questions (MCQs) & 5 Questions with answer as numerical value, with equal weightage to Mathematics, Physics & Chemistry 1st Shift: 09:30 a.m. to 12:30 p.m.

2nd Shift: 02:30 p.m. to 05:30 p.m.

Click Here to predict your JEE Main January 2020 Rank

IMPORTANT DATES FOR JEE Main 2020 – April Exam

JEE Main 2020 Important Dates – April Session

EVENTS

JEE Main 2020 Dates

Application form availability February 7, 2020
Last date to apply March 6, 2020
Last date to upload images and pay application fee March 8, 2020
Issue of admit card March 16, 2020
Exam date April 5 to April 11, 2020
Release date of answer key and recorded responses Second week of April, 2020
Release of answer key Last week of April, 2020: Paper 1
Second week of May, 2020: Paper 2 & 3
Declaration of result April 30, 2020
Announcement of NTA score Third week of May, 2020

 

Eligibility Criteria

There is no age limit for the candidates to appear in JEE Main 2020 Examination. The candidates who have passed their 12th Examination in 2018, 2019 and appearing in 2020 are eligible to JEE Main Examination 2020.

Those candidates who cleared the class 12 exam in 2017 or before 2017 are not eligible to appear for JEE Main 2020 exam.

Educational Qualification

Candidates must have at least 5 subjects in class 12 exam or equivalent exam. Where Math, Physics and Chemistry are the essential subjects.

CLICK HERE TO APPLY FOR JEE MAIN APRIL 2020

INFORMATION BULLETIN for JEE MAIN April 2020

NTA (National Testing Agency) publishes Information Bulletin for JEE MAIN April 2020 in PDF Form. Candidates can download JEE Main 2020 brochure in English and Hindi are available.

The JEE Main 2020 information brochure is that document which provides complete details of the Joint Entrance Examination (Main), officially provided by the exam conducting body NTA. It is recommended to read it first and then apply for the exam.

JEE Main April 2020 Information Bulletin in Hindi DOWNLOAD PDF
JEE Main April 2020 Information Bulletin in English DOWNLOAD PDF

 

JEE Main 2020 Detailed Syllabus

 

If you want to improve %ile for JEE Main April 2020, then Join BOUNCE BACK CRASH COURSE FOR JEE 2020. Register Here!!!

For Details about Bounce Back Crash Course for JEE, Click Here.

 

FREE Revision Series For JEE 2020 | Quick Revision Videos 

👉Physics Revision Series

👉Chemistry Revision Series 

👉Mathematics Revision Series

 

Mind Map For Hydrocarbons: Alkenes | Class 11, JEE & NEET – Download from here

Get to learn all important points and reactions of Hydrocarbons – Alkene through these mind maps. Download and share with your friends also.

Download Mind Map 1

Hydrocarbons Alkene

 

 

Download Mind Map 2

 

 

Download Mind Map 3

Rotational Motion – Part 2 | Download Mind Map for Class 11, JEE & NEET

Rotational Motion Class 11 comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the key formulae and important concepts on finger tips.

Download Mind Map
Rotational Motion part2