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**Previous Years JEE Advanced Questions**

**Column**II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at.M (poteptial at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

(t) Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid-point btween the centres of the rings.

PQ is perpendicular to the line joining the centres and coplanar to the rings.

**[JEE-2009]**

**Sol.**($(\mathrm{A}) \rightarrow \mathrm{p}, \mathrm{r}, \mathrm{s}(\mathrm{B}) \rightarrow \mathrm{r}, \mathrm{s}(\mathrm{C}) \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{t}(\mathrm{D}) \rightarrow \mathrm{r}, \mathrm{s}$)

**Column II**shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown.

**Column I**gives some statements about X and/or Y. Match these statements to the appropriate system(s) from

**Column II**

**[JEE-2009]**

**Sol.**($(A) \rightarrow p, t(B) \rightarrow q, s, t(C) \rightarrow p, r, t(D) \rightarrow q$)

(p)

(A) Total contact force = $\sqrt{\mathrm{N}^{2}+\mathrm{f}^{2}}$ = mg

(B) X is fixed therefore P.E. remain same

(C) Potential energy of ‘Y’ is decreasing as it slide down the plane and K.E. is not increasing

(D) Toque of weight of y is

$\tau=\operatorname{mg} \ell \cos \theta \Rightarrow \tau \neq 0$

(q)

(A) Force due to both ‘X’ and ‘Z’ on ‘Y’ is equal to mg

(B) As the system is moving up P.E. of system is increasing

(C) Mechanical energy of total system is increasing

(D) As the force passes through ‘p’ torque about p is zero

(r)

(A) Force on ‘X’ due to Y

$\mathrm{F}_{\mathrm{net}}=\sqrt{2} \mathrm{T} \Rightarrow \sqrt{2} \mathrm{mg}$

(B) System is moving down therefore P.E. is decreasing

(C) System is moving down with constant speed with constant speed therefore machanical energy is decreasing

(D) $\tau=\operatorname{mg} \ell \Rightarrow \tau \neq 0$

(s)

(A) Force exerted by (X) on ‘Y’ is less than mg and its magnitude $\mathrm{f}_{\mathrm{B}}=\mathrm{Vd}_{\mathrm{t}} \mathrm{g}$

(B) As the ball moves down it displaces the water to its higher position hence P.E. of water increases

(C) As no desssipitative force acts on it mechanical energy is conserved

(D) Torque is non-zero about p

(t) Similar to ‘s’ but it is viscous fluid therefore total mechanical energy decreases with time.

**[2009, 6M]**

**Sol.**7

(A) IBL (B) $\frac{I B L}{\pi}$ (C) $\frac{I B L}{2 \pi}$ $(\mathrm{D}) \frac{I B L}{4 \pi}$

** [2010, 3M]**

**Sol.**(C)

**Comprehension Type**

Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature $\mathrm{T}_{\mathrm{C}}$ (0). An interesting property of superconductors is that their critical temperature becomes smaller than $\mathrm{T}_{\mathrm{C}}$ (0) if they are placed in a magnetic field, i.e., the critical temperature $\mathrm{T}_{\mathrm{C}}$ (B) is a function of the magnetic field strength B. The dependence of $\mathrm{T}_{\mathrm{C}}$ (B) on B is shown in the figure.

**[JEE2010]**

**Sol.**(A)

If $\mathrm{B}_{2}>\mathrm{B}_{1}$, critical temperature, (at which resistance of semiconductors abrupty becomes zero) in case 2 will be less than compared to case 1.

(A) $\mathrm{B}=5$ Tesla, $\mathrm{T}_{\mathrm{C}}(\mathrm{B})=80 \mathrm{K}$

(B) $\mathrm{B}=5$ Tesla, $75 \mathrm{K}<\mathrm{T}_{\mathrm{C}}(\mathrm{B})<100 \mathrm{K}$

(C) $\mathrm{B}=10$ Tesla, $75 \mathrm{K}<\mathrm{T}_{\mathrm{c}}(\mathrm{B})<100 \mathrm{K}$

(D) $\mathrm{B}=10$ Tesla, $\mathrm{T}_{\mathrm{c}}(\mathrm{B})=70 \mathrm{K}$

**[JEE2010]**

**Sol.**(B)

With increase in temperature, TC is decreasing.

$\mathrm{T}_{\mathrm{c}}$ (0) = 100 K

$\mathrm{T}_{\mathrm{c}}$ = 75 K at B = 7.5 T

Hence, at B = 5T, $\mathrm{T}_{\mathrm{c}}$ should line between 75 K and 100 K

(A) they will never come out of the magnetic field region

(B) they will come out travelling along parallel paths

(C) they will come out of the same time

(D) they will come out at different times

**[IIT-JEE 2011]**

**Sol.**(B,D)

By diagram B is true.

*I*’ flows through the wire. The Z-component of the magnetic field at the center of the spiral is

** [IIT-JEE 2011]**

**Sol.**(A)

Taking an elemental strip of radius x and width dx.

Area of strip = $2 \pi \mathrm{x}$ dx

Number of turns through area $=\frac{N}{b-a} d x$

$\int d B=\int_{a}^{b} \frac{\mu_{0} \frac{N}{(b-a)} I d x}{2 x}=\frac{\mu_{0} N I \ell n\left(\frac{b}{a}\right)}{2(b-a)} |$

(A) If $\theta=0^{\circ},$ the charge moves in a circular path in the $\mathrm{x}-\mathrm{z}$ plane.

(B) If $\theta=0^{\circ},$ the charge undergoes helical motion with constant pitch along the y-axis.

(C) If $\theta=10^{\circ},$ the charge undergoes helical motion with its pitch increasing with time, along the

$y$ -axis

(D) If $\theta=90^{\circ},$ the charge undergoes linear but accelerated motion along the $\mathrm{y}$ -axis

**[IIT-JEE 2012]**

**Sol.**(C,D)

2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by $\frac{N}{12} \mu_{0} a J$, then the value of N is

**[IIT-JEE 2012]**

**Sol.**(5)

**[IIT-JEE 2012]**

**Sol.**(D)

(A) The direction of the magnetic field is –z direction.

(B) The direction of the magnetic field is +z direction.

(C) The magnitude of the magnetic field $\frac{50 \pi \mathrm{M}}{3 \mathrm{Q}}$ units.

(D) The magnitude of the magnetic field is $\frac{100 \pi \mathrm{M}}{3 \mathrm{Q}}$ units.

**[IIT-JEE 2013]**

**Sol.**(A,C)

by direction of $\overrightarrow{\mathrm{F}} ;$ magnetic field is in $-\mathrm{z}$ direction

Time $=\frac{\theta}{\omega}=\frac{\pi / 6}{\mathrm{QB} / \mathrm{M}}=\frac{\mathrm{M} \pi}{6 \mathrm{QB}}$

$\Rightarrow \mathrm{B}=\frac{\mathrm{M} \pi}{6 \mathrm{Q}\left(10 \times 10^{-3}\right)}=\frac{50 \pi \mathrm{M}}{3 \mathrm{Q}}$

(A) In the region 0 < r < R, the magnetic field is non-zero

(B) In the region R < r < 2R, the magnetic field is along the common axis

(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centredon the axis

(D) In the region r > 2R, the magnetic field is non-zero

**Sol.**(A,D)

**[JEE-Advanced-2014]**

**Sol.**3

**Paragraph for Questions 15 & 16**

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

(A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h $\approx$ a

(B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h $\approx$ a

(C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h $\approx$ 1.2 a

(D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h $\approx$ 1.2 a

**[JEE-Advanced-2014]**

**Sol.**(C)

$\mathrm{B}_{\text {wire }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \sqrt{\mathrm{a}^{2}+\mathrm{h}^{2}}} 2 \sin \theta$

$=\frac{\mu_{0} \mathrm{ia}}{\pi\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)}$

$\mathrm{B}_{\text {loop }}=\frac{\mu_{0} \mathrm{ia}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)^{3 / 2}}$

$\mathrm{B}_{\text {wire }}=\mathrm{B}_{\text {loop }}$

$\frac{\mu_{0} \mathrm{ia}}{\pi\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)}=\frac{\mu_{0} \mathrm{i} \mathrm{a}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{h}^{2}\right)^{3 / 2}}$

$\sqrt{a^{2}+h^{2}}=\frac{\pi a}{2}$

$a^{2}+h^{2}=2.5 \mathrm{a}^{2}$

$\mathrm{h}^{2}=1.5 \mathrm{a}^{2}$

$\mathrm{h} \approx 1.2 \mathrm{a}$

(A) $\frac{\mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{\mathrm{d}}$

(B) $\frac{\mu_{0} I^{2} a^{2}}{2 d}$

(C) $\frac{\sqrt{3} \mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{\mathrm{d}}$

(D) $\frac{\sqrt{3} \mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{2 \mathrm{d}}$

**[JEE-Advanced-2014]**

**Sol.**(B)

$|\vec{\tau}|=|\vec{M} \times \vec{B}|=M B \sin 30^{\circ}$

where $\mathrm{M}=\left(\pi \mathrm{a}^{2}\right) \mathrm{I}$ and $\mathrm{B}=2 \times \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}=\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{d}}$ Therefore

$\tau=\left(\frac{\mu_{0} \mathrm{I}}{\pi \mathrm{d}}\right)\left(\pi \mathrm{a}^{2} \mathrm{I}\right)\left(\frac{1}{2}\right)=\frac{\mu_{0} \mathrm{I}^{2} \mathrm{a}^{2}}{2 \mathrm{d}}$

(A) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{z}}, \mathrm{F} \propto(\mathrm{L}+\mathrm{R})$

(B) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{x}}, \mathrm{F}=0$

(C) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{y}}, \mathrm{F} \propto(\mathrm{L}+\mathrm{R})$

(D) If $\overrightarrow{\mathrm{B}}$ is along $\hat{\mathrm{z}}, \mathrm{F}=0$

**[JEE-Advanced-2015]**

**Sol.**(A,B,C)

A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity $\overrightarrow{\mathbf{v}}$. A uniform electric field $\overrightarrow{\mathrm{E}}$ and a uniform magnetic field $\overrightarrow{\mathrm{B}}$ exist everywhere. The velocity $\overrightarrow{\mathbf{V}}$, electric field $\overrightarrow{\mathrm{E}}$ and magnetic field $\overrightarrow{\mathrm{B}}$ are given in column 1, 2 and 3, respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude.

(A) (II) (iii) (S) (B) (IV) (i) (S) (C) (III) (ii) (R) (D) (III) (iii) (P)

**[JEE-Advanced-2017]**

**Sol.**(A)

(A) (II) (ii) (R) (B) (IV) (ii) (R) (C) (IV) (i) (S) (D) (III) (iii) (P)

**[JEE-Advanced-2017]**

**Sol.**(C)

For path to be helix with axis along +ve z-direction, particle should experience a centripetal acceleration in x-y plane.

For the given set of options only option (C) satisfy the condition. Path is helical with increasing pitch.

(A) (II) (iii) (S) (B) (IV) (i) (S) (C) (III) (ii) (R) (D) (III) (iii) (P)

**[JEE-Advanced-2017]**

**Sol.**(D)

For particle to move in -ve y-direction, either its velocity must be in –ve y-direction (if initial velocity $\neq$ 0) & force should be parallel to velocity or it must experience a net force in –ve y-direction only (if initial velocity = 0)

**[JEE-Advanced-2017]**

**Sol.**(B)

The given points (1, 2, 3, 4, 5, 6) makes $360^{\circ}$ angle at ‘O’. Hence angle made by vertices 1 & 2 with ‘O’ is $60^{\circ}$.

Direction of magnetic field at ‘O’ due to each segment is same. Since it is symmetric star shape, magnitude will also be same.

Magnetic field due to section BC.

$\left(\mathrm{B}_{1}\right)=\frac{\mathrm{k} \mathrm{i}}{\mathrm{a}}(\sin (+60)-\sin 30)=\frac{\mathrm{ki}}{2 \mathrm{a}}(\sqrt{3}-1)$

$\mathrm{B}_{\mathrm{net}}=12 \times \mathrm{B}_{1}=\frac{6 \mathrm{ki}}{\mathrm{a}}(\sqrt{3}-1) \&\left(\mathrm{k}=\frac{\mu_{0}}{4 \pi}\right)$

(A) For $\mathrm{B}=\frac{8}{13} \frac{\mathrm{p}}{\mathrm{QR}}$, the particle will enter region 3 through the point $P_{2}$ on x-axis

(B) For $\mathrm{B}>\frac{2}{3} \frac{\mathrm{p}}{\mathrm{QR}}$, the particle will re-enter region 1

(C) For a fixed B, particle of same charge Q and same velocity v, the distance between the point $P_{1}$ and the point of re-entry into region 1 is inversely proportional to the mass of the particle.

(D) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the chage in its linear momentum between point $P_{1}$ and the farthest point from y-axis is $\frac{\mathrm{p}}{\sqrt{2}}$.

**[JEE-Advanced-2017]**

**Sol.**(A,B)

For $\mathrm{B}=\fra

c{8}{13} \frac{\mathrm{p}}{\mathrm{QR}},$ radius of path

$\mathrm{R}^{\prime}=\frac{\mathrm{p}}{\mathrm{Q.B}}=\frac{\mathrm{p} \times 13 \mathrm{QR}}{\mathrm{Q} 8 \mathrm{p}}=\frac{13}{8} \mathrm{R}$

using pythagorous theorem, $\mathrm{y}=\frac{5 \mathrm{R}}{8}$

$\therefore$ particle will enter region 3 through point $\mathrm{P}_{2}$

for $\mathrm{B}>\frac{2}{3} \frac{\mathrm{p}}{\mathrm{QR}}$

Radius of path $<\frac{3 \mathrm{PQR}}{\mathrm{Q} 2 \mathrm{p}}=\frac{3}{2} \mathrm{R}$

$\therefore$ Particle will not enter in region $3 \&$ will re-enter region 1

charge in momentum = $\sqrt{2} \mathrm{p}$. When particle enters region 1 between entry point & farthest point from y-axis.

(A) If $\mathbf{I}_{1}$ = $\mathbf{I}_{2}$, then cannot be equal to zero at the origin (0, 0, 0)

(B) If $\mathbf{I}_{1}$ > 0 and $\mathbf{I}_{2}$ < 0, then can be equal to zero at the origin (0, 0, 0)

(C) If $\mathbf{I}_{1}$ < 0 and $\mathbf{I}_{2}$ > 0, then can be equal to zero at the origin (0, 0, 0)

(D) If $\mathbf{I}_{1}$ = $\mathbf{I}_{2}$, then the z-component of the magnetic field at the centre of the loop is $\left(-\frac{\mu_{0} I}{2 R}\right)$

**[JEE-Advanced-2018]**

**Sol.**(A,B,D)

**[JEE-Advanced-2018]**

**Sol.**2

(1) Average speed along x-axis

** [JEE-Advanced-2018]**

**Sol.**5.55