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.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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