An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $24 \mathrm{~cm}^{2}$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is $12 \Omega$. If an externally applied uniform magnetic field in the core along its axis changes from 1.5 T in one direction to 1.5 T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be $\_\_\_\_$ mC.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Electromagnetic Induction

An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area

24 cm^{2}

. The two ends of the wire are connected to a resistor. The total resistance in the circuit is

12Ω

. If an externally applied uniform magnetic field in the core along its axis changes from 1.5 T in one direction to 1.5 T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be

____

mC.

Answer: 60

Solution

The charge flowing through the circuit during a change in magnetic flux is given by

Δ Q = - \frac{Δ ϕ}{R}

.

Initial flux:

ϕ_{1} = N B A = 100 \times 1.5 \times 24 \times 1 0^{- 4}

Wb

Final flux:

ϕ_{2} = - N B A = - 100 \times 1.5 \times 24 \times 1 0^{- 4}

Wb (opposite direction)

Change in flux:

Δ ϕ = ϕ_{2} - ϕ_{1} = - 2 N B A

Therefore:

Δ Q = \frac{2 N B A}{R} = \frac{2 \times 100 \times 1.5 \times 24 \times 1 0 ^{- 4}}{12} = 6 \times 1 0^{- 2} C = 60 mC

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Electromagnetic Induction →

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Electromagnetic Induction

Solution

More JEE Main questions on Electromagnetic Induction