A square loop of side 20 cm and resistance 1Ω is moved towards right with a constant speed v₀. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4Ω. What should be the value of v₀ so that a steady current of 2 mA flows in the loop?

Solution

Physics sir · 3 steps

Only the right arm cuts field lines, so the EMF is Blv₀; then Ohm's law with loop plus network fixes v₀.

1. Find the EMF. Only the right arm, of length l = 0.2 m, is inside the field and cuts across it. The top and bottom arms move along their own length, so they add nothing. The EMF is Blv₀ = 5 × 0.2 × v₀ = v₀ volts, with v₀ in m s⁻¹.

   I = Blv₀/Rₜₒₜₐₗ ⇒ Rₜₒₜₐₗ = v₀/(2 × 10⁻³)

2. Limits on resistance. The current passes through the 1 Ω loop and the network of four 4 Ω resistors. So Rₜₒₜₐₗ cannot be less than 1 Ω and cannot be more than 1 + 16 = 17 Ω.

   Option	v₀ in m s⁻¹	EMF	Rₜₒₜₐₗ needed	Possible?
   (a)	1	1 V	500 Ω	no, too large
   (b)	10⁻²	10⁻² V	5 Ω	yes: 1 Ω loop + 4 Ω network
   (c)	10²	100 V	5 × 10⁴ Ω	no, too large
   (d)	10⁻⁴	10⁻⁴ V	0.05 Ω	no, less than the loop alone

3. So. The network of equal 4 Ω resistors works out to 4 Ω, for example two 8 Ω paths in parallel. With Rₜₒₜₐₗ = 5 Ω we get v₀ = 5 × 2 × 10⁻³ = 10⁻² m s⁻¹, which is 1 cm/s, option (b).

The idea to keep. Only the arm that cuts field lines gives EMF, so ε = Blv. Treat the loop as a battery whose internal resistance is the loop's own, driving the outside network.

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