The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of $1 \mathrm{cms}^{-1}$. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is $1.7 \Omega$, the current in the loop at that instant will be close to :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Electromagnetic Induction

The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of

1 cms^{- 1}

. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is

1.7Ω

, the current in the loop at that instant will be close to :

(A) $115 μ A$
(B) $170 μ A$ ✓ Correct
(C) $60 μ A$
(D) $150 μ A$

Answer: (B)

Solution

Only the edge of the loop that lies in the field sweeps flux, so the induced emf is

ε = B ℓ v = 1 \times 0.05 \times 0.01 = 5 \times 1 0^{- 4} V

. The external resistor network is a balanced Wheatstone bridge, which reduces to an equivalent resistance of

\frac{4}{3} Ω

. Adding the loop's own

1.7 Ω

gives a total of about

3 Ω

, so the current is

i = \frac{ε}{R} = \frac{5 \times 1 0 ^{- 4}}{3} \approx 1.7 \times 1 0^{- 4} A

, close to

170 μ A

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Electromagnetic Induction

Solution

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