The magnetic field in a region is given by B⃗ = Bₒ( x/a )k̂. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity v⃗ = v₀î. The emf induced in the loop is :

Solution

Physics sir · 4 steps

Only the two vertical sides cut field lines, and they sit in different fields, so subtract their EMFs.

1. Which sides matter. The loop moves along x. The top and bottom sides move along their own length, so they give no EMF. Only the two sides parallel to y, each of length d, give a motional EMF Bdv₀.
2. Field at each side. Let the left side be at position x, so the right side is at x + d. The field there is B₀ x/a and B₀ (x+d)/a. Both EMFs drive current the same way around the loop, so they oppose each other and the net EMF is their difference.

   ε = (B₀ (x+d))/a d v₀ - (B₀ x)/a d v₀ = (B₀ v₀ d²)/a

3. Check with flux. The flux is Φ = ∫ₓˣ⁺ᵈ (B₀ x')/a d dx' = (B₀ d)/2a(2xd + d²). Then dΦ/dt = (B₀ d)/2a × 2d v₀ = (B₀ v₀ d²)/a, the same answer, and it does not depend on x.
4. So. The induced EMF is (B₀ v₀ d²)/a, which is option (d).

The idea to keep. For a loop moving through a non-uniform field, find the EMF on the leading and trailing sides separately and take the difference. Only the change in B across the loop matters.

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