Question

A copper wire having a resistance of $0.01\Omega$ per $\text{metre}$ is used to wind a $400$ turn solenoid of radius $1\text{cm}$ and length $20\text{cm}$. Find the strength of the magnetic field inside the solenoid near its centre if emf across the battery is $1\text{V}$.

• A. $10.05\times {10}^{-2}\text{T}$
• B. $1.005\times {10}^{-3}\text{T}$
• C. $1.005\times {10}^2\text{T}$
• D. $1.005\times {10}^{-2}\text{T}$

Answer 1

The correct option is D $1.005\times {10}^{-2}\text{T}$

Given:
$\frac{R}{L}=0.01\Omega /\text{m};r=1\text{cm}=0.01\text{m}$

$N=400\text{turns};E=1\text{V}$

$L=20\text{cm}=0.2\text{m}$

So, net resistance of the solenoid will be

$R=\frac{R}{L}\times 2\pi r\times N$

$R=\left(0.01\Omega /\text{m}\right)(2\pi \times 0.01\text{m})\times 400$

$R=0.25\Omega$

So,
$i=\frac{E}{R}=\frac{1}{0.25}=4\text{A}$

And we know that magnetic field inside the solenoid near its axis is given by

$B={\mu }_{0}ni={\mu }_0\left(\frac{N}{L}\right)i$

$B=4\pi \times {10}^{-7}\times \frac{400}{0.2}\times 4=100.5\times {10}^{-4}$

$\therefore B=1.005\times {10}^{-2}\text{T}$

Hence, option $\left(d\right)$ is the correct answer.