Question

A rectangular coil of length 0.12m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2Weber/m2. The coil carries a current of 2A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be :

Answer 1

The torque required to keep the coil in stable equilibrium will be 0.0012 Nm.

Explanation:

• Initially, the force of the current is calculated by F= current*length*magnetic field=2*0.12*0.2= 0.04 N.
• Now the radius of the wire can be calculated after calculating the area of the coil using the formula, area of the rectangular coil=l*b= 0.12*0.1= 0.012.
• Now the radius of the wire can be calculate as A= (3.14)*radius*radius
• 0.012= 3.14*radius*radius
• radius*radius= (0.012)/3.14= 0.0038, r=0.06.
• Now applying all data in the formula of torque, t= 0.06*0.04*sin(30)= 0.0012 Nm.

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