A copper wire with a resistance of 0.01 ohm per metre is utilized to wind a 400 -turn solenoid having a radius of 1.0 cm and a length of 20 cm . Determine the emf of a battery which, when connected across the solenoid

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Quantrex Academy · Accepted Answer

Correct answer: B. 1.0 V

The total length of the wire used to wind the solenoid is given by:

l = N \times 2 \pi r = 400 \times 2 \pi \times 0.01 = 8 \pi \text{ m}

The total resistance of the solenoid is:

R = l \times \lambda = 8 \pi \times 0.01 = 0.08 \pi \Omega

The magnetic field near the centre of a long solenoid is:

B = \mu_0 n I = \mu_0 \left( \dfrac{N}{L} \right) I

Substituting the given values to find the current

I

:

1.0 \times 10^{-2} = (4 \pi \times 10^{-7}) \times \left( \dfrac{400}{0.2} \right) \times I

1.0 \times 10^{-2} = 8 \pi \times 10^{-4} \times I

I = \dfrac{1.0 \times 10^{-2}}{8 \pi \times 10^{-4}} = \dfrac{100}{8 \pi} = \dfrac{12.5}{\pi} \text{ A}

The emf of the battery required is:

E = I R = \left( \dfrac{12.5}{\pi} \right) \times 0.08 \pi = 1.0 \text{ V}

A copper wire with a resistance of 0.01 ohm per metre is utilized...

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