Question

3. A rectangular coil of wire (a -22.0 cm, b- 43.0 cm) containing a single turn is placed in a uniform 7.80 T magnetic field, as the drawing shows. The current in the loop is 11.0 A. Determine the magnitude of the magnetic force on the top side of the loop.

Answer 1

Electric force in a current carrying wire is given by:

F = i*LxB

F = i*L*B*sin theta

Now theta = 90 deg, for top side of loop, where theta = angle between current and magnetic field

i = current in upper side = 11.0 A

L = length of upper side = a = 22.0 cm = 0.22 m

B = Magnetic field = 7.80 T

So,

F = 11.0*0.22*7.80*sin 90 deg

F = 18.88 N

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