Question

1) Determine the force per meter in a beam of 20,000 A perpendicular to the magnetic field of the Earth (3.00 x 10-5 T).

2) Determine the magnitude of the magnetic field that produces a 2.16 N force on a moving 4.00 cm cable perpendicular to the field carrying a current of 30.0 A.

Answer 1

1)

i = current = 20000 A

B = magnetic field = 3 x 10^-5 T

= 90

magnetic force is given as

F = i B L Sin

F/L = i B Sin90

F/L = (20000) (3 x 10^-5)

F/L = 0.6 N/m

2)

L = 4 cm = 0.04 m

i = current = 30 A

F = magnetic force = 2.16 N

B = magnetic field = ?

= 90

magnetic force is given as

F = i B L Sin

2.16 = (30) B (0.04) Sin90

B = 1.8 T