A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝ tⁿ, then value of n is _________.

Solution

Physics sir · 4 steps

The bar gets longer as it climbs the V, so the EMF grows with time even at constant speed.

1. Where the bar is. The bar starts at the vertex and moves up with constant speed v. After time t it is at height y = vt above the vertex.
2. Length between rails. Let the two rails make angles α and β with the vertical. The part of the bar between the rails then has length l = y(tan α + tan β) = vt(tan α + tan β), so l grows in proportion to t.

   E = Blv = Bv²(tan α + tan β) t

3. Check with flux. The enclosed triangle has area 1/2ly ∝ t², so Φ ∝ t². Then E = dΦ/dt ∝ t, which matches the result above.
4. So. E ∝ t¹, so n = 1.

The common slip: 2. Taking the power of t in the flux, Φ ∝ t², instead of in its derivative gives 2. The EMF is dΦ/dt, which goes as t¹.

The idea to keep. In E = Blv, use only the length of the rod that lies between the rails at that instant. If that length changes with time, the EMF changes too, even when v is constant.

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