Question

a rectangular coil of 25 loops is suspended in a field of 0.20wb/m2.the plane of coil is parallel to the direction of the field the dimensions of the coil are 15cm perpendicular to the field line and 12cm parallel to them what is the current in the coil if there is a torque of 5.4N.M acting on it​

Answer 1

Answer:

The current in the coil is 60 Ampere.

Explanation:

Given:

Number of turns in the coil is N = 25

Dimension of the coil = 15cm X 12cm

magnitude of magnetic field = 0.20T

angle in the xy plane is θ = 0 degree

torque τ = 5.4 N-m

To find:

current in the coil is i = ?

Solution:

The torque acting on the coil is given by

=> $\tau = NiAB cos\theta$

Converting cm to m

12 cm = 0.12 m

15 cm = 0.15 m

The area of the coil is  

A = 0.12 X 0.15

A = $0.018 m^2$

Substituting the values

=>$5.4 = 25\times i \times 0.018 \times 0.20 \times cos\theta$

=>$5.4 = 25\times i \times 0.018 \times 0.20 \times cos(0)$

=>$5.4 = 25\times i \times 0.018 \times 0.20 \times 1$

=>$5.4 = 25\times i \times 0.018 \times 0.20 \times 1$

=>$5.4 = 0.09\times i$

=>$i = \frac{5.4}{0.09}$

=> i = 60 A