A wire carrying a current of 2.40 amps lies between the poles of a magnet. The l

Question

A wire carrying a current of 2.40 amps lies between the poles of a magnet. The length of the wire is 12.0 cm and the magnitude of the uniform magnetic field is 0.38 T

a) What is in Newtons the magnitude of the force on the wire if the magnetic field is parallel to the wire?

b) What is in Newtons the magnitude of the force on the wire if the magnetic field is perpendicular to the wire?

c) What is in Newtons the magnitude of the force on the wire if the angle between the magnetic field and the current is 68.0°?

Explanation / Answer

Here ,

current , I = 2.40 A

L = 0.12 m

magnetic field, B = 0.38 T

a)

for field parallel to wire current ,

theta = 0 degree

force on wire = B*I*L * sin(theta)

as sin(0) = 0

force on wire = 0 N

b)

for perpendicular wire to current ,

theta = 90 degree

force on wire , F = B*I*L * sin(90)

F = 0.38 * 2.4 * 0.12 * 1

F = 0.109 N

force on wire is 0.109 N

c)

for theta = 68 degree

force on wire , F = B*I*L * sin(68)

F = 0.38 * 2.4 * 0.12 * sin(68)

F = 0.101 N

force on wire is 0.101 N

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