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Physics Question on Magnetic Field

A solenoid of length 50 cm and a radius of cross-section 1 cm has 1000 turns of wire wound over it. If the current carried is 5 A, the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space μ0=4π×10−7T−mA−1)\mu _0=4 \pi \times 10^{-7} T-mA^{-1})

A

0.63×10−2T0.63 \times 10^{-2} T

B

1.26×10−2T1.26 \times 10^{-2} T

C

2.51×10−2T2.51 \times 10^{-2} T

D

6.3 T6.3 \, T

Answer

1.26×10−2T1.26 \times 10^{-2} T

Explanation

Solution

The magnetic field is given by \hspace15mm B=\mu _0ni where μ0=4π×10−7T mA−1,\mu _0=4 \pi \times 10^{-7}T \, mA^{-1}, \hspace15mm n=\frac {1000}{50 \times 10^{-2}},i=5 \, A \therefore \hspace15mm B=4 \pi \times 10^{-7} \times \frac {1000}{50 \times 10^{-2}}\times 5 \hspace15mm B=1.26 \times 10^{-2} T