First slide

Magnetism

Question

Two short bar magnets of magnetic moments ‘M' each are arranged at the opposite corners of a square of side ‘d’ such that their corners coincide with the corners and their axes are parallel. If the like pole are in the same direction, the magnetic field at any of the other corners of the square is

Moderate

A

$\frac{μ_{0}}{4 π} . \frac{M}{d^{3}}$
B

$\frac{μ_{0}}{4 π} . \frac{2 M}{d^{3}}$
C

$\frac{μ_{0}}{4 π} . \frac{M}{2 d^{3}}$
D

$\frac{μ_{0}}{4 π} . \frac{M^{2}}{2 d^{3}}$

Solution

Magnetic field at point ‘A’ due to magnet at ‘B’ (axial point) is
$B_{1} = \frac{μ_{0}}{4 π} . \frac{2 M}{d^{3}} (along BA)$
Magnetic field at point ‘A’ due to magnet at ‘D’ ( Equatorial point) is
$B_{2} = \frac{μ_{0}}{4 π} . \frac{M}{d^{3}} (along DA)$

Resultant $B = B_{1} - B_{2}$
$B = \frac{μ_{0}}{4 π} . (\frac{M}{d^{3}})$

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