A short bar magnet of magnetic moment 0.21 A | vedclass

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(D) The magnetic field on the axial line of a short bar magnet is given by $B_{axial} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$.Since the axis of the

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$10$

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(D) The magnetic field on the axial line of a short bar magnet is given by $B_{axial} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$.Since the axis of the magnet is perpendicular to the horizontal component of the earth's magnetic field $(B_H)$,the resultant magnetic field makes an angle of $45^{\circ}$ with $B_H$ when the magnitude of the axial field equals the magnitude of the horizontal component of the earth's magnetic field.Therefore,$B_{axial} = B_H$.Substituting the values: $\frac{10^{-7} 	imes 2 	imes 0.21}{r^3} = 4.2 	imes 10^{-5}$.$r^3 = \frac{2 	imes 0.21 	imes 10^{-7}}{4.2 	imes 10^{-5}} = \frac{0.42 	imes 10^{-7}}{4.2 	imes 10^{-5}} = 0.1 	imes 10^{-2} = 10^{-3} \ m^3$.$r = 0.1 \ m = 10 \ cm$.

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