JEE Advanced PYQ Physics – Motional and Static EMI & Applications of EMI

Q 1 :

A region in the form of an equilateral triangle (in $x - y$ plane) of height $L$ has a uniform magnetic field $\vec{B}$ pointing in the $+ z$ -direction. A conducting loop PQR, in the form of an equilateral triangle of the same height $L$ , is placed in the $x - y$ plane with its vertex P at $x = 0$ in the orientation shown in the figure. At $t = 0$ , the loop starts entering the region of the magnetic field with a uniform velocity $\vec{v}$ along the $+ x$ -direction. The plane of the loop and its orientation remain unchanged throughout its motion.

Which of the following graph best depicts the variation of the induced emf (E) in the loop as a function of the distance $(x)$ starting from $x = 0$ ? [2024]

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(1)

$For x = 0 to L$

$Induced emf, ε = B l_{eff} v = B \times (\frac{x}{\sqrt{3}}) v$

$For x = L to 2 L$

$| emf |$ $= B (\frac{L}{\sqrt{3}} - \frac{x_{0}}{\sqrt{3}}) v - B (\frac{2 x_{0}}{\sqrt{3}}) v$

$= \frac{B v L}{\sqrt{3}} - \sqrt{3} B v x_{0} = B v [\frac{L}{\sqrt{3}} - \sqrt{3} (x - L)]$

$= \frac{B v}{\sqrt{3}} [L - 3 x + 3 L] = \frac{B v}{\sqrt{3}} [4 L - 3 x]$

$At x = \frac{4 L}{3} \Rightarrow emf = 0$

Therefore, graph option (1) best depicts the variation of the induced emf (E) as a function of the distance (x)

Q 2 :

A light disc made of aluminium (a nonmagnetic material) is kept horizontally and is free to rotate about its axis as shown in the figure. A strong magnet is held vertically at a point above the disc away from its axis. On revolving the magnet about the axis of the disc, the disc will (figure is schematic and not drawn to scale) [2020]

rotate in the direction opposite to the direction of magnet's motion
rotate in the same direction as the direction of magnet's motion
not rotate and its temperature will remain unchanged
not rotate but its temperature will slowly rise

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(2)

Due to motion of magnet above the disc, the plate moves through the magnetic flux, due to which an EMF is generated in the plate and eddy currents are induced. These currents are such that it opposes the relative motion so disc will rotate in the same direction as the direction of magnet's motion.

This apparatus is called Arago's disc and the effect was discovered in 1824 by Arago.

Q 3 :

A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of $10 {ms}^{- 1}$ (see figure). [2020]

The e.m.f. induced at the time the left arm of the frame is at $x = 10 cm$ from the wire is:

$2 μ V$
$1 μ V$
$0.75 μ V$
$0.5 μ V$

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(2)

In the given question,

Current flowing through the wire, $I = 1 A$

Speed of the frame, $v = 10 {ms}^{- 1}$

Side of square loop, $l = 10 cm$

Distance of square frame from current carrying wire, $x = 10 cm$

We have to find, e.m.f. induced $e = ?$

According to Biot-Savart's law,

$B = \frac{μ_{0} I d l \sin θ}{4 π x^{2}} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{1 \times 10^{- 1}}{{(10^{- 1})}^{2}} = 10^{- 6}$

Induced e.m.f., $e = B l v = 10^{- 6} \times 10^{- 1} \times 10 = 1 μ V$

Q 4 :

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be [2001]

maximum in situation (a)
maximum in situation (b)
maximum in situation (c)
the same in all situations

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(1)

When current flows in any of the coils, the flux linked with the other coil is maximum when surface area to receive flux is maximum. Clearly the flux linkage is maximum in case (1) due to the spatial arrangement of the two circular coils.

Q 5 :

A metallic square loop ABCD is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced [2001]

in AD, but not in BC
in BC, but not in AD
neither in AD nor in BC
in both AD and BC

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(4)

Electric field will be induced, as ABCD moves, in both AD and BC. The metallic square loop moves in its own plane with velocity $v$ in a uniform magnetic field perpendicular to the plane of the square loop. AD and BC are perpendicular to the velocity as well as perpendicular to the applied magnetic field.

Q 6 :

A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges parallel to the X and Y axes. The region $y \geq 0$ has a uniform magnetic field, $\vec{B} = B_{0} \hat{k} .$ The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts to enter the magnetic field with an initial velocity $v_{0} \hat{j} m/s,$ as shown in the figure. Considering the quantity $K = \frac{B_{0}^{2} L^{2}}{R M}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct: [2025]

If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field.
When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) \ln (\frac{5}{2}) .$
If $v_{0} = 3 K L$ , the complete loop enters inside the region of magnetic field at time $t = (\frac{1}{K}) \ln (\frac{3}{2}) .$

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Select one or more options

(2, 4)

$Current i = \frac{ε}{R} = \frac{B v ℓ}{R}$ $[∵ ε = B v ℓ]$

$\vec{F} = B (\hat{i}) (ℓ) (- \hat{j})$

$M a = - B_{0} [\frac{B_{0} v ℓ}{R}] (ℓ)$ $\Rightarrow a = - \frac{B_{0}^{2} ℓ^{2} v}{M R}$

$Given K = \frac{B_{0}^{2} ℓ^{2} V}{R M}$ $∴$ $a = - K v$

$\frac{d v}{d t} = - k v$ or, $\int_{v_{0}}^{v} \frac{d v}{d t} = \int_{0}^{t} (- k) d t$ $\Rightarrow \ln (\frac{v}{v_{0}}) = - k t$

$v = v_{0} e^{- k t}$ ...(i)

$\frac{d x}{d t} = v_{0} e^{- k t}$ $(∵ x \leq l)$

$\int_{0}^{x} d x = \int_{0}^{t} v_{0} e^{- k t} d t$ $= \frac{v_{0}}{k} (1 - e^{- k t})$

When $x = L$

$L = \frac{v_{0}}{k} (1 - e^{- k t})$

If $v_{0} = 3 k L$

$L = \frac{3 k L}{k} (1 - e^{- k t})$ $\Rightarrow \frac{1}{3} = 1 - e^{- k t}$

or $\frac{2}{3} = e^{- k t}$ $∴$ $t = \frac{1}{k} \ln (\frac{3}{2})$

Complete loop will enter at $t = \frac{1}{k} \ln (\frac{3}{2})$

So, option (4) is correct.

And as the loop is completely inside, then further no current flows and no force acts, i.e., $F = 0$

So option (2) is correct.

Q 7 :

A conducting wire of parabolic shape, initially $y = x^{2}$ is moving with velocity $\vec{V} = V_{0} \hat{i}$ in a non-uniform magnetic field $\vec{B} = B_{0} (1 + {(\frac{y}{L})}^{β}) \hat{k},$ as shown in the figure. If $V_{0} B_{0}$ , $L$ and $β$ are positive constants and $Δ ϕ$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are: [2019]

$| Δ ϕ |$ is proportional to the length of the wire projected on the $y$ -axis.
$| Δ ϕ |$ remains the same if the parabolic wire is replaced by a straight wire, $y = x$ initially, of length $\sqrt{2} L$ .
$| Δ ϕ |$ $= \frac{1}{2} B_{0} V_{0} L for β = 0$
$| Δ ϕ |$ $= \frac{4}{3} B_{0} V_{0} L for β = 2$

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Select one or more options

(1, 2, 4)

Given: $B = B_{0} [1 + {(\frac{y}{L})}^{β}], \vec{V} = V_{0} \hat{i}$

We now consider an infinitesimally small length of wire $d y$ at a distance $y$ from the origin.

Emf induced across the length dy $| ∆ ϕ |$ $= B (d y) V_{0}$

$| ∆ ϕ |$ $= B_{0} [1 + {(\frac{y}{L})}^{β}] V_{0} d y$

$∴$ Induced emf across the complete projection

$| Δ ϕ |$ $= B_{0} V_{0} \int_{0}^{L} [1 + {(\frac{y}{L})}^{β}] V_{0} d y$ $= B_{0} V_{0} L [1 + \frac{1}{β + 1}]$

For $β = 0$ , $| Δ ϕ |$ $= 2 B_{0} V_{0} L$

Clearly, $| Δ ϕ |$ $\propto L$

For $β = 2$ , $| Δ ϕ |$ $= \frac{4}{3} B_{0} V_{0} L$

For a straight wire of length $\sqrt{2} L$ , placed along $y = x$ , the value of $| Δ ϕ |$ will remain the same as its projection on the $y$ -axis is the same $L$ as that of the previous.

Q 8 :

A rigid wire loop of square shape having side of length L and resistance R is moving along the $x$ -axis with a constant velocity $v_{0}$ in the plane of the paper. At $t = 0$ , the right edge of the loop enters a region of length 3L where there is a uniform magnetic field $B_{0}$ into the plane of the paper, as shown in the figure. For sufficiently large $v_{0}$ , the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v (x)$ , $I (x)$ and $F (x)$ represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of $x$ . Counter-clockwise current is taken as positive. [2016]

Which of the following schematic plot(s) is (are) correct?

(Ignore gravity)

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Select one or more options

(1, 2)

$i = \frac{e}{R} = \frac{B L v}{R}$ ...(i)

[Counter-clockwise direction while entering, zero when completely inside and clockwise while exiting]

$F = i L B = \frac{B^{2} L^{2} v}{R}$ ...(ii)

[Toward left while entering and exiting and zero when completely inside]

$∴$ $- m v \frac{d v}{d x} = \frac{B^{2} L^{2} v}{R}$

$∴$ $\int_{v_{0}}^{v} d V = - \frac{B^{2} L^{2}}{m R} \int_{0}^{x} d x$ $\Rightarrow V - V_{0} = - \frac{B^{2} L^{2}}{m R} x$

$∴$ $V = V_{0} - \frac{B^{2} L^{2} x}{m R}$ ...(iii)

[V decreases from $x = 0$ to $x = L$ , remains constant for $x = L$ to $x = 3 L$ , again decreases from $x = 3 L$ to $x = 4 L$ hence graph (1) is correct]

From (i) and (iii),

$i = \frac{B L}{R} [V_{0} - \frac{B^{2} L^{2} x}{m R}]$

[ $i$ decreases from $x = 0$ to $x = L$ , $i$ becomes zero from $x = L$ to $x = 3 L$ , $i$ changes direction and decreases from $x = 3 L$ to $x = 4 L$ ]

Hence graph (2) is correct.

Q 9 :

A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the $90 °$ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of $10 {A s}^{- 1} .$ Which of the following statement(s) is(are) true? [2016]

The magnitude of induced $e m f$ in the wire is $(\frac{μ_{0}}{π})$ volt
If the loop is rotated at a constant angular speed about the wire, an additional $e m f$ of $(\frac{μ_{0}}{π})$ volt is induced in the wire.
The induced current in the wire is in opposite direction to the current along the hypotenuse.
There is a repulsive force between the wire and the loop.

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Select one or more options

(1, 4)

The flux passing through the triangular wire if $i$ current flows through the infinitely long conducting wire

$d ϕ = \int_{0}^{0.1} \frac{μ_{0} i}{2 π x} \times 2 π d x$

$ϕ = \frac{μ_{0} i}{10 π} = M i$

$∴$ $M = \frac{μ_{0}}{10 π}$ $(\frac{d i}{d t} = 10 {As}^{- 1} given)$

Induced emf in the wire, $e = M \frac{d i}{d t} = \frac{μ_{0}}{10 π} \times 10 = \frac{μ_{0}}{π} V$

There will be no extra induced emf in the wire because there is no change in the magnetic flux due to rotation of the loop.

As the current in the triangular wire is decreasing, the induced current in AB is in the same direction as the current in the hypotenuse of the triangular wire. Therefore, force will be repulsive.

Q 10 :

A thin conducting rod MN of mass 20 gm, length 25 cm and resistance $10 Ω$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_{0} = 4 T$ directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t = 0$ and it moves down along the rails. Assume air drag is negligible.

Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.

[Given: The acceleration due to gravity $g = 10 {ms}^{- 2}$ and $e^{- 1} = 0.4$ ] [2023]

List-I List-II

(P) At t = 0.2 s, the magnitude of the induced emf in volt (1) 0.07

(Q) At t = 0.2 s, the magnitude of the magnetic force in newton (2) 0.14

(R) At t = 0.2 s, the power dissipated as heat in watt (3) 1.20

(S) The magnitude of terminal velocity of the rod in ${ms}^{- 1}$ (4) 0.12

(5) 2.00

P → 5, Q → 2, R → 3, S → 1
P → 3, Q → 1, R → 4, S → 5
P → 4, Q → 3, R → 1, S → 2
P → 3, Q → 4, R → 2, S → 5

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(4)

From force equation, $m g - B i ℓ = m a$

$= m g - B i ℓ = m \frac{d v}{d t}$ $\Rightarrow m g - B (\frac{B i ℓ}{R}) ℓ = m \frac{d v}{d t}$ $[∵ i = \frac{ε}{R} = \frac{B ℓ v}{R}]$

$\Rightarrow \frac{m g R}{B^{2} ℓ^{2}} - v = \frac{m R}{B^{2} ℓ^{2}} \frac{d v}{d t}$

$\frac{B^{2} ℓ^{2}}{m R} \int_{0}^{t} d t = \int_{0}^{v} \frac{d v}{\frac{m g R}{B^{2} ℓ^{2}} - v}$

or, $\frac{B^{2} ℓ^{2}}{m R} = \frac{16 \times \frac{1}{16}}{20 \times 10^{- 3} \times 10} = \frac{1}{0.2} = 5$

Now, $\frac{m g R}{B^{2} ℓ^{2}} = \frac{20 \times 10^{- 3} \times 10 \times 10}{16 \times \frac{1}{16}} = 2$

And, $\frac{B^{2} ℓ^{2}}{m R} = \frac{16 \times \frac{1}{16}}{20 \times 10^{- 3} \times 10} = \frac{1}{0.2} = 5$

$∴$ $5 t = {[- \ln (2 - v)]}_{0}^{v}$ $\Rightarrow - 5 t = \ln (\frac{2 - v}{v})$

$∴$ $v = 2 (1 - e^{- 5 t})$

At $t = 0.2 s$ ,

$v = 2 (1 - e^{- 5 \times 0.2})$

$= 2 (1 - 0.4)$

$= 1.2 m/s$

At $t = 0.2 s$ , induced emf $ε = B v ℓ$

$∴$ $ε = 4 \times 1.2 \times \frac{1}{4} = 1.2 V$

Magnetic force $= B i ℓ \sin θ = B \times \frac{B ℓ v}{R} \times ℓ \times \sin 90 °$

$= \frac{4 \times 4 \times \frac{1}{4} \times 1.2 \times \frac{1}{4}}{10} = 0.12 N$

Power dissipated as heat, $P = i^{2} R - \frac{V^{2}}{R}$

$∴$ $P = \frac{1.2 \times 1.2}{10} = 0.144 W$

At terminal velocity, the net force becomes zero,

$∴$ $m g = B i ℓ$ $\Rightarrow m g = B \times \frac{B ℓ v_{t}}{R} \times ℓ$

$∴$ $v_{T} = \frac{m g R}{B^{2} ℓ^{2}} = \frac{20 \times 10^{- 3} \times 10 \times 10}{16 \times \frac{1}{16}} = 2 m/s$

Topic Question Set

Q 4 :

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be [2001]

Q 5 :

A metallic square loop ABCD is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced [2001]

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	List-I		List-II
(P)	At t = 0.2 s, the magnitude of the induced emf in volt	(1)	0.07
(Q)	At t = 0.2 s, the magnitude of the magnetic force in newton	(2)	0.14
(R)	At t = 0.2 s, the power dissipated as heat in watt	(3)	1.20
(S)	The magnitude of terminal velocity of the rod in ${ms}^{- 1}$	(4)	0.12
		(5)	2.00

Topic Question Set

Q 4 : Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be [2001]

Q 5 : A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced [2001]

Want Us to Email you About Special Offers & Updates?

Q 4 :

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be [2001]

Q 5 :

A metallic square loop ABCD is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced [2001]