Moving Charges And Magnetism Ques 1
1. A long solenoid of $50 \mathrm{cm}$ length having $100$ turns carries a current of $2.5 \mathrm{A}$. The magnetic field at the centre of the solenoid is :
[2020]
$\left(\mu_0=4 \pi \times 10^{-7} \mathrm{T} \mathrm{m} \mathrm{A}^{-1}\right)$
(a) $3.14 \times 10^{-4} \mathrm{T}$
(b) $6.28 \times 10^{-5} \mathrm{T}$
(c) $3.14 \times 10^{-5} \mathrm{T}$
(d) $6.28 \times 10^{-4} \mathrm{T}$
Show Answer
Answer:
Correct Answer: 1.(d)
Solution: (d) Magnetic field at the centre of solenoid,
$ B_{\text {solenoid }}=\mu_0 n l $
Given : No. of turns / length,
$ n=\frac{N}{L}=\frac{100}{50 \times 10^{-2}}=200 \text { turns } / \mathrm{m} $
Current, $I=2.5 \mathrm{A}$
$ \begin{aligned} \therefore B_{\text {solenoid }} & =\mu_0 n I=4 \pi \times 10^{-7} \times 200 \times 2.5 \\ & =6.28 \times 10^{-4} \mathrm{T} \end{aligned} $
BETA
