LEVELJEE Main

Animated Solution for Physics - Electromagnetic Induction: As shown in the figure, $P$ and $Q$ are two coaxial conducting loops separated by some distance. When the switch $S$ is closed, a clockwise current $I_{p}$ flows in $P$ (as seen by $E$ ) and an induced current $I_{Q_{1}}$ flows in $Q$ . The switch remains closed for a long time. When $S$ is opened, a current $I_{Q_{2}}$ flows in $Q$ . Then the direction $I_{Q_{1}}$ and $I_{Q_{2}}$ (as seen by $E$ ) are

Select Answer:

Visualized Solution

$S e t u p$

Two coaxial loops $P$ and $Q$ .
Observer $E$ is on the right.

$L e n z^{'} s L a w$

Induced current opposes the change in magnetic flux.

$C l os in g S w i t c h S$

Current $I_{p}$ flows clockwise as seen by $E$ .

$M a g n e t i c F i e l d B_{P}$

By right-hand rule, $B_{P}$ points from right to left.

$I n d u ce d C u r r e n t I_{Q_{1}}$

Leftward flux increases.
To oppose, $Q$ produces rightward field.
Requires anti-clockwise current $I_{Q_{1}}$ .

$O p e nin g S w i t c h S$

Current $I_{p}$ decreases.
Leftward flux decreases.

$I n d u ce d C u r r e n t I_{Q_{2}}$

To oppose decrease, $Q$ produces leftward field.
Requires clockwise current $I_{Q_{2}}$ .

$C o n c l u s i o n$

$I_{Q_{1}}$ is anti-clockwise.
$I_{Q_{2}}$ is clockwise.

00:00 / 00:00

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Solution Diagram

Understanding Lenz's Law with Coaxial Loops

Imagine you are standing at position

E

, looking directly at two coaxial conducting loops,

P

and

Q

. This problem is a classic and beautiful demonstration of Lenz's Law and the Right-Hand Thumb Rule.

Analyzing the Closing of the Switch

When the switch

S

is closed, a current

I_{p}

begins to flow in loop

P

. The problem explicitly states that, from your perspective at

E

, this current is clockwise.

Let's apply the Right-Hand Thumb Rule. If you curl the fingers of your right hand in a clockwise direction, your thumb points directly away from you. This means the magnetic field

B_{P}

produced by loop

P

points from right to left, passing through loop

Q

As the current

I_{p}

builds up from zero to its steady value, this leftward magnetic flux through

Q

is increasing. According to Lenz's Law, the induced current in

Q

must oppose this change. To fight the increasing leftward flux, loop

Q

must generate its own magnetic field pointing to the right (towards you at

E

Using the Right-Hand Thumb Rule in reverse, to create a magnetic field pointing towards you, the induced current

I_{Q_{1}}

must flow in an anti-clockwise direction.

Analyzing the Opening of the Switch

Now, let's look at what happens when the switch

S

is opened after being closed for a long time. The steady current

I_{p}

in loop

P

begins to drop rapidly to zero.

Consequently, the leftward magnetic flux through loop

Q

starts to decrease. Lenz's Law dictates that loop

Q

will again oppose this change. This time, it doesn't want the leftward flux to disappear! To support the dying flux, loop

Q

will induce a current that produces its own magnetic field pointing to the left (away from you).

To generate a leftward magnetic field, the induced current

I_{Q_{2}}

must flow in a clockwise direction.

Final Conclusion

By carefully tracking the changes in magnetic flux and applying Lenz's Law, we have determined that the initial induced current

I_{Q_{1}}

is anti-clockwise, and the subsequent induced current

I_{Q_{2}}

is clockwise. This perfectly matches option (d).

Common Traps & FAQs

Why does the magnetic field point from right to left?

When viewed from E (on the right), the current in P is clockwise. Using the right-hand thumb rule, curling your fingers clockwise makes your thumb point away from you, which is towards the left.

What happens when the switch remains closed for a long time?

The current in P becomes steady, meaning the magnetic flux through Q is constant. Since there is no change in flux, the induced current in Q becomes zero.

Why is

I_{Q_{1}}

anti-clockwise?

Closing the switch increases the leftward magnetic flux through Q. To oppose this increase, Q must create a rightward magnetic field. By the right-hand rule, a rightward field requires an anti-clockwise current as seen from E.

Why does opening the switch cause a clockwise current?

Opening the switch decreases the leftward flux. To oppose this decrease, Q tries to maintain the leftward field by producing its own leftward magnetic field, which requires a clockwise current.

Does the distance between the loops affect the direction of the induced current?

No, the distance only affects the magnitude of the induced current, not its direction. The direction is solely determined by whether the flux is increasing or decreasing.

Similar Questions

JEE Advanced 1989

LEVELJEE Main

A conducting square loop of side $L$ and resistance $R$ moves in its plane with a uniform velocity $v$ perpendicular to one of its sides. A magnetic induction $B$ , constant in time and space, pointing perpendicular to and into the plane of the loop exists everywhere. The current induced in the loop is

(A)

B Lv / R

clockwise

(B)

B Lv / R

anti-clockwise

(C)

2 B Lv / R

A very long solenoid of radius $R$ is carrying current $I (t) = k t e^{- α t} (k > 0)$ , as a function of time $(t \geq 0)$ . Counter clockwise current is taken to be positive. A circular conducting coil of radius $2 R$ is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by

A region in the form of an equilateral triangle (in $x - y$ plane) of height $L$ has a uniform magnetic field $B$ 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 $v$ along the $+ x$ -direction. The plane of the loop and its orientation remain unchanged throughout its motion.

A coil of wire having finite inductance and resistance has a conducting ring placed co-axially within it. The coil is connected to a battery at time $t = 0$ so that a time dependent current $I_{1} (t)$ starts flowing through the coil. If $I_{2} (t)$ is the current induced in the ring and $B (t)$ is the magnetic field at the axis of the coil due to $I_{1} (t)$ , then as a function of time ( $t > 0$ ), the product $I_{2} (t) B (t)$

does not vary with time

(D)

passes through a maximum

JEE Advanced 2017

LEVELJEE Advanced

A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2 A$ as shown in the figure. At the point of crossing, the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $B$ points into the plane of the paper. At $t = 0$ , the loop starts rotating about the common diameter as axis with a constant angular velocity $ω$ in the magnetic field. Which of the following options is/are correct?

* Multiple Correct Options

(A)

The emf induced in the loop is proportional to the sum of the areas of the two loops.

(B)

The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper.

(C)

The net emf induced due to both the loops is proportional to

cos ω t

(D)

The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

JEE Advanced 1985

LEVELJEE Advanced

Space is divided by the line $A D$ into two regions. Region I is field free and the region II has a uniform magnetic field $B$ directed into the plane of the paper. $A C D$ is a semicircular conducting loop of radius $r$ with centre at $O$ , the plane of the loop being in the plane of the paper. The loop is now made to rotate with a constant angular velocity $ω$ about an axis passing through $O$ and perpendicular to the plane of the paper. The effective resistance of the loop is $R$ . (a) Obtain an expression for the magnitude of the induced current in the loop. (b) Show the direction of the current when the loop is entering into the region II. (c) Plot a graph between the induced current and the time of rotation for two periods of rotation.

LEVELBoard

An infinitely long cylinder is kept parallel to a uniform magnetic field $B$ directed along positive Z-axis. The direction of induced current as seen from the Z-axis will be

(A)

clockwise of the +ve Z-axis

(B)

anti-clockwise of the +ve Z-axis

(C)

zero

(D)

along the magnetic field

JEE Main 2025

LEVELJEE Advanced

A conducting square loop initially lies in the $X Z$ plane with its lower edge hinged along the $X$ -axis. Only in the region $y \geq 0$ , there is a time dependent magnetic field pointing along the $Z$ -direction, $B (t) = B_{0} (cos ω t) \hat{K}$ , where $B_{0}$ is a constant. The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts rotating with constant angular speed $ω$ about the $X$ axis in the clockwise direction as viewed from the $+ X$ axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ( $V$ ) in the loop as a function of time:

A small circular loop of area $A$ and resistance $R$ is fixed on a horizontal xy-plane with the center of the loop always on the axis $\overset{n}{^}$ of a long solenoid. The solenoid has $m$ turns per unit length and carries current $I$ counterclockwise as shown in the figure. The magnetic field due to the solenoid is in $\overset{n}{^}$ direction. List-I gives time dependences of $\overset{n}{^}$ in terms of a constant angular frequency $ω$ . List-II gives the torques experienced by the circular loop at time $t = \frac{π}{6 ω}$ , Let $α = \frac{A ^{2} μ _{0}^{2} m ^{2} I ^{2} ω}{2 R}$ .

List-I

(P)

\frac{1}{2} (sin ω t \hat{j} + cos ω t \hat{k})

(Q)

\frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{j})

(R)

\frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{k})

(S)

\frac{1}{2} (cos ω t \hat{i} + sin ω t \hat{k})

List-II

(1)

0

(2)

- \frac{α}{4} \hat{i}

(3)

\frac{3 α}{4} \hat{i}

(4)

\frac{α}{4} \hat{j}

(5)

- \frac{3 α}{4} \hat{i}

JEE Main 2004

LEVELJEE Main

A coil having $n$ turns and resistance $R Ω$ is connected with a galvanometer of resistance $4 R Ω$ . This combination is moved for time $t$ seconds from a magnetic field $W_{1}$ weber to $W_{2}$ weber. The induced current in the circuit is

(A)

- \frac{W _{2} - W _{1}}{5 R n t}

(B)

- \frac{n ( W _{2} - W _{1} )}{5 R t}

(C)

- \frac{( W _{2} - W _{1} )}{R n t}

(D)

- \frac{n ( W _{2} - W _{1} )}{R t}

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lenz′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Closing Switch S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Field BP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced Current IQ1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Opening Switch S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced Current IQ2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Understanding Lenz's Law with Coaxial Loops

Analyzing the Closing of the Switch

Analyzing the Opening of the Switch

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along positive Z-axis. The direction of induced current as seen from the Z-axis will be

List-I

List-II

$S e t u p$

$L e n z^{'} s L a w$

$C l os in g S w i t c h S$

$M a g n e t i c F i e l d B_{P}$

$I n d u ce d C u r r e n t I_{Q_{1}}$

$O p e nin g S w i t c h S$

$I n d u ce d C u r r e n t I_{Q_{2}}$

$C o n c l u s i o n$

An infinitely long cylinder is kept parallel to a uniform magnetic field $B$ directed along positive Z-axis. The direction of induced current as seen from the Z-axis will be