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 B⃗ 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 :

Solution

Physics sir · 3 steps

The bar splits the rails into two loops: one shrinks, one grows, so their currents circulate opposite ways.

1. Two separate loops. The bar, the rails and R₁ form the left loop. The bar, the rails and R₂ form the right loop. Moving the bar left makes the left loop smaller and the right loop bigger.

   Loop	Area	Flux into page	Induced field must point	Current
   Left (R₁)	shrinks	decreases	into page, to make up the loss	clockwise
   Right (R₂)	grows	increases	out of page, to oppose the gain	anticlockwise

2. Check with the bar. On a positive charge in the bar, qv⃗ × B⃗ points down, so current flows down the bar. In the left loop, down the bar on its right side is clockwise. In the right loop, down the bar on its left side is anticlockwise.
3. So. I₁ is clockwise and I₂ is anticlockwise, which is option (b).

The idea to keep. When a sliding bar is shared by two loops, apply Lenz's law to each loop on its own. The loop losing flux and the loop gaining flux always get opposite senses of current.

Answer from the NTA final answer key. Solution by PrepWiser.