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  • Radius and number of turns of a circular coil are 10cm and 25 respectively. What should be the current through the coil that will produces a magnetic field of <img alt="6.28 \times 10^{-5} \ma

Radius and number of turns of a circular coil are 10cm and 25 respectively. What should be the current through the coil that will produces a magnetic field of 6.28 \times 10^{-5} \mathrm{wb} / \mathrm{m}^2 at its centre is 

Option: 1

0.3A


Option: 2

0.2A


Option: 3

0.4A


Option: 4

0.12A


Answers (1)

best_answer

Radius = \frac{10}{2}=5 \mathrm{~cm}=0.05 \mathrm{~m}

no. of turns = N = 25

\begin{aligned} & \therefore B=\frac{\mu_0 N I}{2 r} \\ & I=\frac{B 2 r}{\mu_0 \ N}=\frac{2 \times 0.05 \times 6.28 \times 10^{-5}}{4 \times \pi \times 10^{-7} \times 25} \\ & =0.2 \mathrm{~A} \\ & \end{aligned}

 

Posted by

Ritika Harsh

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