Question

A long solenoid of \(50 \mathrm{~cm}\) length having 100 turns carries a current of \(2.5 A\). The magnetic field at the centre of the solenoid is : \(\left(\mu_{o}=4 \pi \times 10^{-7} T m A^{-1}\right)\)
(A) \(6.28 \times 10^{-4} T\)
в \(3.14 \times 10^{-4} T\)
c \(6.28 \times 10^{-5} T\)
D \(3.14 \times 10^{-5} \mathrm{~T}\)

Answer 1

Solution:
Magnetic field at centre of solenoid \(=\mu_{0} n I\)
$$
\begin{aligned}
&n=\frac{N}{L} \\
&=\frac{100}{50 \times 10^{-2}} \\
&=200 \mathrm{tums} / \mathrm{m} \\
&I=2.5 \mathrm{~A}
\end{aligned}
$$
On putting the values,
$$
\begin{aligned}
&B=4 \pi \times 10^{-7} \times 200 \times 2.5 \\
&=6.28 \times 10^{-4} T
\end{aligned}
$$

Answer 2

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Answer 3

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