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Question

Physics Question on Magnetic properties of materials

Two short magnets each of dipole moment M, are fastened perpendicularly at their centres. The magnitude of magnetic field at a distance d from the centre on the bisector of the right angle is

A

μ04π\frac{\mu_0}{4\pi} 2M2d3\frac{2M\sqrt{2}}{d^3}

B

μ04π\frac{\mu_0}{4\pi} Md3\frac{M}{d^3}

C

μ04π\frac{\mu_0}{4\pi} M2d3\frac{M\sqrt{2}}{d^3}

D

μ04π\frac{\mu_0}{4\pi} 2Md3\frac{2M}{d^3}

Answer

\frac{\mu_0}{4\pi}$$\frac{2M\sqrt{2}}{d^3}

Explanation

Solution

Resultant magnetic moment of the two magnets is MR=M2+M2=M2M_R = \sqrt{M^2 + M^2} = M \sqrt{2} Imagine a short magnet lying along OP with magnetic moment equal to MRi.e.M2M_R \, i.e. \, M\sqrt{2} Thus point P lies on the axial line of the magnet \therefore Magnitude of magnetic field at P is given by, B = μ04π2M2d3\frac{\mu_0}{4 \pi} \frac{2M \sqrt{2}}{d^3} (Baxial=μ04π2Mr3)\left( \because \, B_{axial} = \frac{\mu_0}{4 \pi} \frac{2M}{r^3} \right)