Question

If a current of $0.8\text{A}$ is flowing through a solenoid of $10000\text{turns}$ having a total length of $30\text{cm}$, then what is the magnitude of the magnetic field?

• A. $3.53\times {10}^{-2}\text{T}$
• B. $3.35\times {10}^2\text{T}$
• C. $3.35\times {10}^{-2}\text{T}$
• D. $4.35\times {10}^{-2}\text{T}$

Answer 1

The correct option is C $3.35\times {10}^{-2}\text{T}$

Given,
$i=0.8\text{A}$

$n=\frac{N}{L}=\frac{10000}{30\times {10}^{-2}}=\frac{10000}{0.3}$

$\therefore B={\mu }_{0}ni$

$=4\pi \times {10}^{-7}\times \frac{10000}{0.3}\times 0.8$

$=3.35\times {10}^{-2}\text{T}$

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