A square coil of side 10 cm consists of 20 turns and carries a currentof 12 A. The coil is suspended vertically and the normal to the planeof the coil makes an angle of 30º with the direction of a uniformhorizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Given: side of the square coil is 10 cm , number of turns is 20 , the current carried by the coil is 12 A , angle between the plane of the coil and the magnetic field is 30 ° , and magnitude of the magnetic field is 0.80 T .
Area of square coil is given as,
A = l × l = 10 × 10 cm 2 × ( 1 m 2 10 4 cm 2 ) = 0.01 m 2
Torque experienced by the coil is given as,
τ = n B I A sin θ
where, the magnetic field is B , current carried in the wire is I , area of the square coil is A , angle between the plane of the coil and magnetic field is θ , and side of the square is l .
By substituting the given values in the above equation, we get,
τ = 20 × 0.8 × 12 × 0.01 × sin 30 0 = 0.96 Nm
Therefore, torque experienced by the coil is 0.96 Nm .
Given: side of the square coil is
Area of square coil is given as,
Torque experienced by the coil is given as,
where, the magnetic field is
By substituting the given values in the above equation, we get,
Therefore, torque experienced by the coil is
