Question

A copper wire having resistance 0.01 ohm in each metre is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0 × 10⁻² T near the centre of the solenoid.

Answer 1

Given:
Resistance per unit length of the wire, $\frac{R}{l}$ = 0.01 Ω/m
Radius of the wire, r = 1.0 cm = 0.01 m
Total no. of turns, N = 400
Magnetic field intensity, B = 1.0 × 10⁻² T
Now,
Let E be the emf of the battery and R₀ be the total resistance of the wire.
$$\begin{gathered} \therefore i=\frac{E}{R_0}=\frac{E}{0.01\times 2\mathrm{\pi }r\times 400} \\ =\frac{E}{0.01\times 2\times \mathrm{\pi }\times 0.01\times 400} \end{gathered}$$

The magnetic field near the centre of the solenoid is given by
$$\begin{gathered} B={\mathrm{\mu }}_{0}ni \\ =1\times {10}^{-2}=4\mathrm{\pi }\times {10}^{-7}\times \frac{400}{20\times {10}^{-2}}\times \frac{E}{2\mathrm{\pi }\times 4\times {10}^{-2}} \\ \Rightarrow E=\frac{{10}^{-2}\times 20\times {10}^{-2}\times 2\times {10}^{-2}}{{10}^{-7}\times 4\times {10}^2} \\ =1\mathrm{V} \end{gathered}$$