A short magnetic needle is placed in a magnetic field B i in the direction ( 3 i + j ) . The needle experiences a torque of 0.06 ~N - m . If the same magnetic needle is placed in a magnetic field 2 B j in the direction ( i + 3 j ) , the torque experienced by it is

Step-by-step solution

( ) Torque on a magnet (dipole moment m ) when it is placed in region of magnetic field (Intensity B) is = M B Where, M = M . u M= magnitude of dipole moment and u = unit vector in direction of M Now in case I given, B =B i and M = M 2 ( 3 i + j ) Hence, torque is ₁= M 2 ( 3 i + j ) B i = M B 2 ( j i )= M B 2 (- k ) Now, given, | ₁ |=0.06 ~N - m So, 0.06=M B / 2 or M B=0.12 units Now, when M =M ( i + 3 j 2 ) and B =2 B j then, torque will be ₂= M B = M 2 ( i + 3 j ) 2 B j =M B( i j )=M B( k ) Magnitude of torque is