Question

The magnetic moment of a short bar magnet is 2Am2. The magnetic field at a point 1m away from it in a direction making an angle 60 degree with its magnetic moment is

Answer 1

Magnetic field due to a bar magnet of magnetic dipole moment M at a distance r from it in a direction θ with its magnetic moment is given by

B = (μ0/(4π)). (M/r3). (3cos2θ + 1)½.

μ0/(4π) = 10−7 T.m/A,

M = 2 A m2, 

r = 1 m

θ = 60°

Substituting all the above values in the expression for B, we get

B = (10−7 T.m/A) x (2 A m2 / 1 m3 ) (3/4 + 1)½.

   = √7  x  10−7  T.

Answer 2

Answer:

The magnetic field at a point 1m distance in a direction 60°  with magnetic moment is $\sqrt{7}\times10^{-7} T$ .

Explanation:

Given the magnetic moment of a short bar magnet, $M= 2A.m^{2}$

         distance of a point, $r= 1m$

         angle between magnetic field and magnetic moment is 60°

The net magnetic field due to a magnetic dipole moment M at a distance r from it in a direction θ with its magnetic moment is

                                  $B =\frac{\mu_{0} }{4\pi } \frac{M}{r^{3} } \sqrt{ 3cos^{2} \theta+1 }$

Substituting, $\frac{\mu_{0} }{4\pi } =10^{-7} \frac{T.m}{A}$

        magnetic field, $B =10^{-7}\times \frac{2}{1^{3} } \times \sqrt{ 3cos^{2} 60^{o} +1 }$

                                      $=10^{-7}\times 2\times \sqrt{ 3(\frac{1}{2} )^{2} +1 }$

                                      $=10^{-7}\times 2\times \sqrt{ \frac{7}{4} }$

                                      $=10^{-7}\times 2\times\frac{\sqrt{7} }{2} }=\sqrt{7}\times10^{-7} T$

Hence, the magnetic field due to a magnetic dipole moment at 1m is   $\sqrt{7}\times10^{-7} T$