Question

A coil of length 4cm and width 2cm and 45 turns is placed in a uniform field of flux density 0.8T. if the axis of the coil is at right angle to d magnetic field and has a current of 12A, calculate the turning moment on the coil

Answer 1

Answer

Area of coil

A=$Length \times width$

=$0.04\times0.02=0.0008m^2$

No. Of turns

$N=45$

Mangnetic field density

B=0.85T

Right angle means

$\theta=90$

Therefore

the turning moment on the coil

$\tau=NIABsin\theta$

$=45\times12\times0.0008\times0.85\times sin90°$

=390.15 N-m