Question

A circular coil consits of N=14 loops each of diameter 1.0 m. the coil is placed in a external magnetic field of 0.5T. The angle between the normal to the plane of the coil and the magnitic field is 45 degrees. Calculate the torque in Nm in exerted by the magnetic field on the coil when i=6A

Answer 1

Given data:-

No of turns of circular loop(N) = 14

Diameter of a circular loop (d) = 1.0

Magnetic filed (B) = 0.5 T

Current flowing through coil (i) = 6 Amps

Angle between the normal to the plane and B = 45 deg

• Formulae ;- Torque exerted by the magnetic field(T) = BiNASin
• Area of the circular coil (A) = r^2
• Area = 3.14*(1.0/2)^2 = 0.785 m^2

​​​​​​​substitute all the values in a above formulae ;

we get ;

T = 0.5*6*14*0.785*sin(45)

T=​​​​​​​ 32.97* sin(45) ( where sin45 =0.707)

T = 23.313 N-m

Hence Torque exerted by the magnetic field (T) =23.313 N-m