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Question

The magnetic field at a point inside a 2 mH inductor coil becomes 0.8 of its maximum value in 20 μs when the inductor is joined to a battery. Find the resistance of the circuit.
[Take ln(0.2)=1.609]

A
144 Ω
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B
127 Ω
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C
133 Ω
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D
161 Ω
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Solution

The correct option is D 161 Ω
Let the resistance of the circuit is R and the time constant of the circuit is τ.

At any time, t current through the LR circuit is,

i=i0(1et/τ) .......(1)

As we know, the magnetic field inside a long coil (solenoid) is,

B=μ0ni ...........(2)

From (1) and (2) we get,

B=μ0ni0(1et/τ)

Assuming, B0=μ0ni0

B=B0(1et/τ) ........(3)

Here,
B=0.8B0 ; L=2 mH ; t=20 μs

Putting this values in (3) we get,

0.8B0=B0(1e20×106/τ)

e20×106/τ=0.2

Taking log on both sides we get,

20×106τ=ln (0.2)

τ=20×1061.609 [ln (0.2)=1.609]

The time constant of LR circuit is,

τ=LR=20×1061.609

2×103R=20×1061.609

R=160.9 Ω161 Ω

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer.

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