Ratio of magnetic field at the center of a current carrying circular loop of radius ‘\( R \)’ with that at a distance of ‘\( 4R \)’ from the center on its axis is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 17\sqrt{17} \)
The field on the axis of a circular loop at distance \( x \) from the centre is
\( \displaystyle B_x = \frac{\mu_0IR^{2}}{2(R^{2}+x^{2})^{3/2}}, \qquad B_{\mathrm{centre}} = \frac{\mu_0I}{2R} \)
Dividing:
\( \displaystyle \frac{B_{\mathrm{centre}}}{B_x} = \frac{(R^{2}+x^{2})^{3/2}}{R^{3}} = \left(1 + \frac{x^{2}}{R^{2}}\right)^{3/2} \)
Putting \( x = 4R \):
\( \displaystyle \frac{B_{\mathrm{centre}}}{B_x} = (1 + 16)^{3/2} = 17^{3/2} = 17\sqrt{17} \)