JEE Main 2021 — Electromagnetic Induction Question with Solution

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 $$\Omega$$

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 $$\times$$ 10^-3 A

Induced emf, e = Bv₀I

Induced current, I = e / R_eq

Substituting the values in the above equation, we get

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

$$\Rightarrow$$ v₀ = 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.