Question

A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.

Answer 1

The induced current in the loop at the given dimension is 10.25 A.

The given parameters:

• number of turns of the solenoid, n = 1200 turns
• diameter of the solenoid, d = 2.0 cm
• length of the solenoid, L = 7.5 cm
• resistance of the loop, R = 0.032 Ω.
• current in solenoid, I = 1.3 A
• time, t = 30 ms

The area are of the solenoid is calculated as;

$A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.02^2}{4} = 0.000314 \ m^2$

The emf induced in the solenoid is calculated as;

$emf = N\frac{d\phi}{dt} \\\\emf = N (\frac{BA}{t} )\\\\emf = N(\frac{\mu_0 NI \times A} {L\times t} )\\\\emf = \frac{N^2 \mu_0 I A}{Lt} \\\\emf = \frac{(1200)^2 \times (4\pi \times 10^{-7}) \times (1.3) \times (0.000314)}{0.075 \times 30\times 10^{-3}} \\\\emf = 0.328 \ V$

The induced current in the loop is calculated as follows;

$I = \frac{emf}{R} \\\\I = \frac{0.328}{0.032} \\\\I = 10.25 \ A$

Thus, the induced current in the loop is 10.25 A.

"Your question is not complete, it seems to be missing the following information;"

A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.If the current in the solenoid is increased by 1.3 A in 30 ms, what is the induced current in the loop?

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