The magnetic field B due to a current carrying circular loop of radius 12 cm at its centre is 0.5 × 10−4 T . The magnetic field due to this loop at a point on its axis at a distance of 5.0 cm from its centre would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3.9 × 10−5 T
Both fields come from the Biot–Savart law for a circular loop:
\( \displaystyle B_{\mathrm{centre}} = \frac{\mu_0 I}{2R} \qquad B_{\mathrm{axis}} = \frac{\mu_0 I R^{2}}{2(R^{2} + x^{2})^{3/2}} \)
The current is not given, so divide the two expressions; the current cancels and only the geometry ratio remains.
\( \displaystyle \frac{B_{\mathrm{axis}}}{B_{\mathrm{centre}}} = \frac{R^{3}}{(R^{2} + x^{2})^{3/2}} \)
With R = 12 cm and x = 5.0 cm (the units cancel in the ratio):
\( \displaystyle R^{2} + x^{2} = 144 + 25 = 169 \;\Rightarrow\; (R^{2} + x^{2})^{3/2} = 13^{3} = 2197 \)
\( \displaystyle B_{\mathrm{axis}} = 0.5\times10^{-4}\,\mathrm{T} \times \frac{1728}{2197} \)
\( \displaystyle B_{\mathrm{axis}} \approx 0.393\times10^{-4}\,\mathrm{T} \approx 3.9\times10^{-5}\,\mathrm{T} \)