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 are shown in figure (b).

This pair of forces produce 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 – eSaral
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 the middle finger points in the direction of the magnetic field then the thumb points towards the direction of the force or motion of the conductor.

Then if the forefinger 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 the force.

(ii) Right-hand Palm Rule: Stretch the fingers and thumb of the 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 an 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

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

So, that’s all from this blog. I hope you get the idea about the Flemings Left Hand Rule of Class 12. If you enjoyed this article then please share it with your friends.

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Force on a Current-Carrying Conductor in 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 a magnitude equal to the product of current with a small part of the 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 the 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 solenoids and Toroids and the differences between both.

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-conducting material.
Let a toroid having N closely spaced turns of wire,
the magnetic field in the region occupied by the torus = B
Radius of the Toroid ring = R
To calculate the field, we must evaluate $\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}$ 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 toroid 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 a 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 the 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 the 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:

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 the current is passed through them, they both act as 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 use 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 cardboard or china-clay. When an electric current is passed through the solenoid, a magnetic field is produced around and within the solenoid.

The 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 indicates 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 the cross-section of the solenoid. Just as a parallel-plate capacitor produces a 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 that helps us to calculate the magnetic field, And yes, 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,

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 a 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 the z-axis. Let I will 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 the 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   Click here for the Video tutorials of Magnetic Effect of Current Class 12

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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 Centre 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 center 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 the 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 center.

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 ZZ’ [dB $\cos \alpha$ and $\left.d B^{\prime} \cos \alpha\right]$ 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 the magnetic field is only along PX will contribute to the magnetic field due to the whole loop at point P.

The magnetic dipole moment of the current loop

The current loop can be regarded as a magnetic dipole that produces its magnetic field and the magnetic dipole moment of the current loop is equal to the product of ampere-turns and area of the 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
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Helmholtz Coils | Magnetic Field between two Coils | eSaral

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 a distance equal to the radius of either of the coils and in which the same current in the 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 the middle point rate of change of magnetic field is constant, so if distance increases from a coil magnetic field decrease, but the distance from 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 a magnetic field at the center of a single coil.)\

Also Read: Biot Savart’s Law

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Right Hand Palm Rule – Magnetic Effect of Current Class12, JEE & NEET
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 to 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 the 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 the magnetic field near the center 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 points 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

Hey, do you want to learn about the Magnetic Field due to Infinite Straight Conductor? If yes. Then keep reading.

Magnetic field due to long straight conductor

Here we will discuss all the cases involved in the 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 that OP makes with the direction of the 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 a 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 the 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 an 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 an electric current. The 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. Oersted discovered a magnetic field around a conductor carrying an 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 produces 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 addition 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.

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

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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Wheatstone Bridge Circuit Diagram, Applications, Formula – Class 12, JEE & NEET
The Wheatstone Bridge circuit is nothing more than two simple series-parallel arrangements of resistances connected between a voltage supply terminal and ground producing zero voltage difference between the two parallel branches when balanced. Here we will study about, Wheatstone Bridge, the Conditions of balanced and Unbalanced Wheatstone bridge circuit Diagram, and its applications.

It is an arrangement of four resistances devised by Charles Wheatstone which is used to measure an unknown resistance.

Principle of Wheatstone Bridge

The Wheatstone bridge principle states that if four resistances P, Q, R, and S are arranged to form a bridge with a cell and key between A and C, and a galvanometer between B and D then the bridge is said to be balanced when the galvanometer shows a zero deflection.

In balanced condition$\mathrm{I}_{9}=0 \quad$ so $\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{D}}$ or $\frac{P}{Q}=\frac{R}{s}$. This is called condition of balance.

Balanced Wheatstone Bridge

The condition of balance depends on resistance P, Q, R, and S. This is independent of the emf of the battery.

In-state of balance, the cell and galvanometer can be interchanged.   While performing an experiment at start press the cell key $K_{c}$ first and then the galvanometer key $\mathrm{K}_{9}$ and at end remove $\mathrm{K}_{9}$ first and then $\mathrm{K}_{\mathrm{c}}$ to avoid induced effects.

P and Q are called ratio arms, R is the known resistance arm and S is the unknown arm. BD and AC are called conjugate arms.

The resistance of a balanced Wheatstone bridge between A and C is $\frac{(P+Q)(R+S)}{P+Q+R+S}$.

The sensitivity of the bridge depends upon the value of resistances. The sensitivity is maximum when P, Q, R, and S are of the same order.

Unbalanced Wheatstone bridge

If $\frac{P}{Q}<\frac{R}{s}$ then $V_{B}>V_{D}$ so current flows from $B$ to $D .$

If $\frac{P}{Q}>\frac{R}{s}$ then $V_{B}<V_{D}$ so current flows from $D$ to $B$

Applications of Wheatstone Bridge

The Wheatstone Bridge has many uses in electronic circuits other than comparing an unknown resistance with a known resistance. When used with Operational Amplifiers, the Wheatstone bridge circuit can be used to measure and amplify small changes in resistance, RX due.

Wheatstone bridge is not suitable for measurement of very small and very high resistances. Very low resistances are measured by Kelvin’s double bridge while very high resistances by leakage method.

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Mind Map for Modern Physics: Radioactivity Revision – Class XII, JEE, NEET
Radioactivity in Modern Physics comprises variety of cases with important formulae and key points. So here is the mind map to help you in remembering all the formulas and important key concepts on finger tips.

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eSaral provides detailed Notes of Physics, Chemistry, Mathematics and Biology Notes for class 11, and 12. So here you will get class 11 notes for mathematics. There are important points in Mathematics such as formulae, equations, identities, properties, theorem etc. what has to be remembered to solve problems in Math.eSaral is providing complete study material to prepare for IIT JEE, NEET and Boards Examination. So here Trigonometry Height and Distance Notes with Important questions for IIT JEE Exam preparation. With the help of Notes, candidates can plan their Strategy for a particular weaker section of the subject and study hard. So, go ahead and check the Important Notes for CBSE Class 11 Maths.

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eSaral provides detailed Notes of Physics, Chemistry, Mathematics and Biology Notes for class 11, and 12. So here you will get class 12 notes for mathematics. There are important points in Mathematics such as formulae, equations, identities, properties, theorem etc. what has to be remembered to solve problems in Math.eSaral is providing complete study material to prepare for IIT JEE, NEET and Boards Examination. Here are the detailed Inverse Trigonometric Function Problems that will help in IIT JEE and boards preparation. With the help of Notes, candidates can plan their Strategy for a particular weaker section of the subject and study hard. So, go ahead and check the Important Notes for CBSE Class 12 Maths. These Notes have been designed in the most simple and precise format covering almost all the Domains like Differential Calculus, Algebra, Trigonometry, and Coordinate Geometry. These Notes will also be helpful to those who are preparing for competitive Exams like JEE.

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eSaral provides detailed Notes of Physics, Chemistry, Mathematics and Biology Notes for class 11, and 12. So here you will get class 11 notes for mathematics. There are important points in Mathematics such as formulae, equations, identities, properties, theorem etc. what has to be remembered to solve problems in Math.eSaral is providing complete study material to prepare for IIT JEE, NEET and Boards Examination. So here trigonometry class 11 notes for IIT JEE Exam preparation. With the help of Notes, candidates can plan their Strategy for a particular weaker section of the subject and study hard. So, go ahead and check the Important Notes for CBSE Class 11 Maths.   eSaral helps the students by providing you an easy way to understand concepts and attractive study material for IIT JEE which includes the video lectures & Study Material designed by expert IITian Faculties of KOTA. eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE preparation. We hope these Mathematics Notes for Class 11 will be helpful for you to understand the important topics and help in remembering the key points for the exam point of view. Get Complete Class 11 Mathematics Notes  For free video lectures and complete Course material for Class 10, JEE and NEET Download eSaral APP.   About eSaral At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone. eSaral Making Learning affordable, Accessible, and thoughtful for all.
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