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  • Radius and number of turns of a circular coil are 10 cm and 25 respectively. What should be the current through the coil that will produces a magnetic field of <img alt="6.28\times10^{-5}wb/m^

Radius and number of turns of a circular coil are 10 cm and 25 respectively. What should be the current through the coil that will produces a magnetic field of 6.28\times10^{-5}wb/m^2 at it center is 

Option: 1

0.3 A


Option: 2

0.2 A


Option: 3

0.4 A


Option: 4

0.12 A


Answers (1)

best_answer

Radius =\frac{10}{2}= 5 cm =0.05m

No. of turns = N = 25

\therefore B =\frac{\mu_0NI}{2A}

I = \frac{B2A}{\mu_0N}=\frac{2\times0.05\times6.2}{4\times \pi \times 10^{-9}} = 0.2A

Posted by

Pankaj

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