A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?​

6.8 N.m

Explanation:

The computation of the magnitude of the magnetic torque on the coil is given below:

Given that

n = 400

d =  6.0 cm

Current  is I = 7.0 A

Angle is = 30 degree

Now

We know that

the magnitude of the magnetic torque is

= nIABsin

= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)

As

= (90° - Ф)

=  (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°

= 6.8 N.m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west