Magnetic field due to current element

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Question

A 200 -turn solenoid having a length of 5 m and a diameter of 10 cm carries a current of 0.30 A. The magnitude of the magnetic field $\vec{B}$ inside the solenoid is $1.5 \times 10^{- p} T$ . Find the value of $p$ .

Easy

Solution

The magnetic field at the center of the solenoid is given by
$B = μ_{0} n i = \frac{4 π \times 10^{- 7} \times 200 \times 0.3}{5} = 1.5 \times 10^{- 5} T$

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