Question

A $50\, cm$ long solenoid has winding of $400$ turns. What current must pass through it to produce a magnetic field of induction $4 \pi \times 10^{-3} \, T$ at the center?

• A. $10.5\, A$
• B. $12.5 \, A$
• C. $25.0 \, A$
• D. $20.0 \, A$

Answer 1

Answer: (B)

For a solenoid,
$B=\mu_{0} n I $
$\Rightarrow n=\frac{N}{L}$
Here, $B=4 \pi \times 10^{-3} T$
$n=\frac{400}{50 \times 10^{-2}}$
$=\frac{400 \times 100}{50}=800$ turns $/ m$
Since, $B=\mu_{0} n I$
$4 \pi \times 10^{-3}=4 \pi \times 10^{-7} \times 800 \times I$
$\Rightarrow I=\frac{4 \pi \times 10^{-3}}{4 \pi \times 10^{-7} \times 800} $
$\Rightarrow I=\frac{10000}{800}=12.5 A$