Question

A square loop of side 20 cm and resistance 1: 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 4\(\Omega\) What should be the value of v₀ so that a steady current of 2 mA flows in the loop ?

(1) 1 m/s

(2) 1 cm/s

(3) 10² m/s

(4) 10^-2 cm/s

Answer 1

Correct Option is (2) 1 cm/s

According to given circuit diagram, equivalent resistance between point P and Q.

\(R _{PQ} = (4+4)|| (4+4)\)

\(=\frac {8\times 8}{8 + 8} = 4 \Omega\)

The equivalent circuit can be drawn as,

Equivalent resistance, R_eq = 4 + 1 = 5 Ω

Magnetic field, B = 5T

The side of the square loop, I = 20 cm = 0.20 m

The steady value of the current, I = 2 mA = 2 × 10^-3 A

Induced emf, e = Bv₀I

Induced current, I = e / R_eq

Substituting the values in the above equation, we get

\(2\times 10 ^{-3} = \frac {5 \times v_0\times 0.2}{5}\)

⇒ \(v_0 = 10 ^{-2} m/s = 1 \,cm /s\)

\(\therefore\) The value of v₀ = 1 cm/s, so that a steady current of 2 mA flows in the loop.

Answer 2

Correct Option is (2) 1 cm/s

Equivalent circuit