Two identical short bar magnets each of magnetic moment \( 12.5\,\mathrm{A\,m^{2}} \) are placed with 10 cm separation between their centers as shown in the figure. The magnitude of magnetic field at O is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{5} \times 10^{-2}\,\mathrm{T} \)
O is 5 cm from the centre of each magnet. It lies on the axial line of magnet P and on the equatorial line of magnet Q, whose axis is perpendicular to that of P. For a short magnet:
\( \displaystyle B_{\mathrm{axial}} = \frac{\mu_0}{4\pi}\,\frac{2M}{d^{3}}, \qquad B_{\mathrm{equatorial}} = \frac{\mu_0}{4\pi}\,\frac{M}{d^{3}} \)
\( M = 12.5\,\mathrm{A\,m^{2}} \)
\( d = 5\,\mathrm{cm} = 0.05\,\mathrm{m},\ d^{3} = 1.25\times10^{-4}\,\mathrm{m^{3}} \)
\( \mu_0/4\pi = 10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle B_1 = \frac{10^{-7} \times 2 \times 12.5}{1.25\times10^{-4}} = 2\times10^{-2}\,\mathrm{T} \quad \text{(along the axis of P)} \)
\( \displaystyle B_2 = \frac{10^{-7} \times 12.5}{1.25\times10^{-4}} = 1\times10^{-2}\,\mathrm{T} \quad \text{(parallel to the axis of Q)} \)
The two fields are perpendicular, so they add as vectors:
\( \displaystyle B = \sqrt{B_1^{2} + B_2^{2}} = \sqrt{2^{2} + 1^{2}} \times 10^{-2} = \sqrt{5}\times10^{-2}\,\mathrm{T} \)