A galvanometer coil has 500 turns and each turn has an average area 3\times 10^{-4}m^{2}. If a torque of 1.5\; Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 \; A is flowing through it, the strength of the field (in T) is _______.
Option: 1 20
Option: 2 10
Option: 3 22
Option: 4 30

Answers (1)

As coil is parallel to magnetic field, the Area vector or Magnetic Moment is perpendicular to magnetic field i.e., \theta= 90^{0}

Torque applied due to magnetic field when angle between magnetic moment and magnetic field \theta is

\tau = MB\sin \theta= NIAB\sin \theta

\\N=500 turns \\A=3\times 10^{-4}m^{2} \\\tau_{external} = 1.5Nm \\I=0.5A

\\\tau =1.5= 500\times 0.5\times 3\times 10^{-4}B\sin \90^{0} \\B=20 T

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