Question

A $50cm$ 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. $10.5A$
• B. $12.5A$
• C. $25.0A$
• D. $20.0A$

Answer 1

Answer: (B)

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