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 (in volt) which when connected across the solenoid will cause a magnetic field of $1 .0 \times 10^{- 2}$ T near the centre of the solenoid.

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Detailed Solution

$\begin{array}{l} Resistance per unit length = \frac{R}{l} = 0 .01 Ω m^{- 1}, \\ r = 1 .0 cm Total turns = 400, ℓ = 20 cm, \\ B = 1 \times 10^{- 2} T, n = \frac{400}{20 \times 10^{- 2}} turns / m \\ i = \frac{E}{R_{0}} = \frac{E}{\frac{R}{l} \times (2 πr \times 400)} = \frac{E}{0 .01 \times 2 \times π \times 0 .01 \times 400} \\ B = μ_{0} ni \\ \Rightarrow 10^{- 2} = 4 π \times 10^{- 7} \times \frac{400}{20 \times 10^{- 2}} \times \frac{E}{400 \times 2 π \times 0 .01 \times 10^{- 2}} \end{array}$
$\Rightarrow E = \frac{10^{- 2} \times 20 \times 10^{- 2} \times 400 \times 2 π \times 10^{- 2} \times 0 .01}{4 π \times 10^{- 7} \times 400} = 1 V$