Question

A 200-turn rectangular coil having dimensions of 3.0 cm by 6.0 cm is placed in a uniform magnetic field of magnitude 0.76 T. (a) Find the current in the coil if the maximum torque exerted on it by the magnetic field is 0.14 N · m. A (b) Find the magnitude of the torque on the coil when the magnetic field makes an angle of 25° with the normal to the plane of the coil. N · m

Answer 1

Answer:

(a). Current, I = 0.512 A = 512 mA

(b). Torque, $\tau = 0.059 N.m$

Given:

Number of turns in the rectangular coil, n = 200 turns

Area of the coil with dimensions 3.0 cm by 6.0 cm, A = $3.0\times 6.0 = 18.0 cm^{2} = 18.0\times 10^{- 4} m^{2}$

Intensity of magnetic field, B = 0.76 T

maximum torque, $\tau_{max} = 0.14 N.m.A$

angle, $\theta = 25^{\circ}$

Solution:

(a) Current in the coil, I can be calculated by the given relation:

$\tau_{max} = nABI$

Therefore,

$I = \frac{\tau_{max}}{nAB}$

Now substituting the given values in the above eqn:

$I = \frac{0.14}{200\times 18.0\times 10^{-4} \times 0.76}$

I = 0.512 A = 512 mA

(b)  Magnitude of torque can be calculated by the given relation:

$\tau = \tau_{max}sin\theta$

Now,

$\tau = 0.14\times sin25^{\circ}$

$\tau = 0.059 N.m$