A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed v . The correct statement about the directions of induced currents $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ flowing through $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ respectively is:

(D) $\mathrm{I}_{1}$ is in clockwise direction and $\mathrm{I}_{2}$ is in anticlockwise direction

(A) $\mathrm{I}_{1}$ is in anticlockwise direction and $\mathrm{I}_{2}$ is in clockwise direction

(B) Both $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ are in anticlockwise direction

(C) Both $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ are in clockwise direction

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Electromagnetic Induction

A conducting bar of length

L

is free to slide on two parallel conducting rails as shown in the figure. Two resistors

R_{1}

and

R_{2}

are connected across the ends of the rails. There is a uniform magnetic field

B

pointing into the page. An external agent pulls the bar to the left at a constant speed v .

The correct statement about the directions of induced currents

I_{1}

and

I_{2}

flowing through

R_{1}

and

R_{2}

respectively is:

(A) $I_{1}$ is in anticlockwise direction and $I_{2}$ is in clockwise direction
(B)Both $I_{1}$ and $I_{2}$ are in anticlockwise direction
(C)Both $I_{1}$ and $I_{2}$ are in clockwise direction
(D) $I_{1}$ is in clockwise direction and $I_{2}$ is in anticlockwise direction✓ Correct

Answer: (D)

Solution

As the bar is pulled to the left, the loop on the left formed by the bar, the rails and

R_{1}

shrinks, so the into-page flux threading it falls. The induced current opposes that fall by maintaining into-page flux inside this loop, which requires it to circulate clockwise.
At the same moment the loop on the right formed by the bar, the rails and

R_{2}

grows, so its into-page flux rises. The induced current there opposes the rise by setting up out-of-page flux inside that loop, which requires it to circulate anticlockwise.
So

I_{1}

flows clockwise and

I_{2}

flows anticlockwise.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Electromagnetic Induction

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