A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure Two…

A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure

Two resistors R1 and R2 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 I1 and I2 flowing through R1 and R2 respectively is :

  1. Both I1 and I2 are in anticlockwise direction
  2. Both I1 and I2 are in clockwise direction
  3. I1 is in clockwise direction and I2 is in anticlockwise direction
  4. I1 is in anticlockwise direction and I2 is in clockwise direction

Solution

Given, the external agent pulls the bar towards left. According to Lenz's law and Fleming's right hand rule, there will be a motional EMF generated in the bar. The right-hand rule tells us that the magnetic force on positive charges will be in downward direction and the electric equivalent circuit can be drawn as shown below.

As velocity is in the right direction, flux through left loop increases and in right loop decreases. According to Lenz's law, current will be induced in that direction where it will resist the change in flux. So in left loop, it will be in anti clockwise direction and in right its in clock wise direction which will help to increase magnetic flux in right side of coil

Asked in: JEE Main 2021 (16 Mar Shift 1)

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