Question

What torque acts on a 40 turn coil of 100 cm² area carrying a current of 10 ampere held with its axis at right angles magnetic field of flux density 0-2 tesla and plane of the col parallel to the field?​

Answer 1

Torque will be :

$\sf \tau = M \times B \times \sin( {90}^{ \circ} )$

• 'M' is magnetic moment, 'B' is field intensity.

• $\theta$ = 90° , because the plane of the coil is parallel to the direction of magnetic field.

$\sf \implies \tau = (n \times i \times area)\times B \times \sin( {90}^{ \circ} )$

$\sf \implies \tau = (40 \times 10 \times \dfrac{100}{10000} )\times 0.2\times 1$

$\sf \implies \tau = 4\times 0.2$

$\sf \implies \tau = 0.8 \: Nm$

Hope It Helps.