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The magnetic field at the centre of a circular coil of radius r and carrying a current I is B. what is the magnetic field at a distance \mathrm{x}=\sqrt{3} \mathrm{r} from the centre, on the axis of the coil ?

Option: 1

B


Option: 2

2 B


Option: 3

4 B


Option: 4

8 B


Answers (1)

best_answer

The magnetic field at a distance \mathrm{x=\sqrt{3} r} is
                      \mathrm{ B^{\prime}=\frac{\mu_0 ~I ~n ~r^2}{2\left(r^2+x^2\right)^{3 / 2}}=\frac{\mu_0 ~I ~n ~r^2}{2\left(r^2+3 r^2\right)^{3 / 2}} \\ }

                            \mathrm{ =\frac{\mu_0 ~I ~n}{16 r} \\ }


But \mathrm{ \quad B=\frac{\mu_0 ~I ~n}{2 r} }


\mathrm{ \therefore B^{\prime}=8 B }.

Hence the correct choice is (4).

Posted by

HARSH KANKARIA

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