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Physics-

General

Easy

Question

Two short magnets of equal dipole moments M are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is

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

The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 2 square root of 2 M over denominator d to the power of 3 end exponent end fraction$

Resultant magnetic moment of the two magnets is
$M subscript n e t end subscript equals square root of M to the power of 2 end exponent plus M to the power of 2 end exponent end root equals square root of 2 M$

Imagine a short magnet lying along OP with magnetic moment equal to $M square root of 2$ . Thus point P lies on the axial line of the magnet.
$therefore$ Magnitude of magnetic field at P is given by $B equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 M over denominator d to the power of 3 end exponent end fraction$

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