Two short magnet of moments M1 and M2 are kept mutually perpendicular to each other with their centres coniciding - Physics - Magnetism And Matter

Question

Two short magnet of moments M1 and M2 are kept mutually
perpendicular to each other with their centres
coniciding The resultant field at a point on the axis of M1 is
inclined at a 300 with axis of M1 The ratio of
M1 to M2 is
(A) 1: 2 (B) 3 : 4 (C) 3 : 2 (D) 2 : 3

Kranav Sharma · Accepted Answer

$\vec{B_{1}} = \frac{µ_{0} M_{1}}{2 π r^{3}}, \vec{B_{2}} = \frac{µ_{0} M_{2}}{4 π r^{3}} Resultant of \vec{B_{1}}, and \vec{B_{2}}, | \vec{B_{}} | = \sqrt{\vec{{B_{1}^{2}}_{}} + \vec{{B_{2}^{2}}_{}}} = \frac{µ_{0}}{4 π r^{3}} \sqrt{4 M_{1} r^{2} + M_{2}^{2}} resolving \vec{B_{}} into its components$

$B_{x} = | \vec{B_{}} | \cos 60 ° = | \vec{B_{}} | \frac{1}{2} B_{y} = - | \vec{B_{}} | \sin 60 ° = - | \vec{B_{}} | \frac{1}{2}$

Move over,

$\frac{\vec{B_{1}}}{\vec{B_{2}}} = \frac{M_{1}}{2} \times \frac{4}{M_{2}} = \frac{2 M_{1}}{M_{2}} Or, \frac{M_{1}}{M_{2}} = \frac{\vec{B_{1}}}{2 \vec{B_{2}}} Or, M_{1} : M_{2} = \vec{B_{1}} : 2 \vec{B_{2}}$

Two short magnet of moments M1 and M2 are kept mutually perpendicular to each other with their centres coniciding. The resultant field at a point on the axis of M1 is inclined at a 300 with axis of M1. The ratio of M1 to M2 is (A) 1: 2 (B) 3 : 4 (C) 3 : 2 (D) 2 : 3

Two short magnet of moments M1 and M2 are kept mutually perpendicular to each other with their centres

coniciding. The resultant field at a point on the axis of M1 is inclined at a 300 with axis of M1. The ratio of

M1 to M2 is

(A) 1: 2 (B) 3 : 4 (C) 3 : 2 (D) 2 : 3