Question

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.15-N force on the 4.00 cm of wire in the field. What is the average field strenth?

Answer 1

Answer:

1.79 T

Explanation:

Applying,

F = BILsin∅................ Equation 1

Where F = Force, B = magnetic field, I = current flowing through the wire, L = length of the wire, ∅ = angle between the magntic field and the force

make B the subject of the equation

B = F/ILsin∅............. Equation 2

From the question,

Given: F = 2.15 N, I = 30 A, L = 4.00 cm = 0.04 m, ∅ = 90° (perpendicular to the field)

Substitute these values into equation 2

B = 2.15/(30×0.04×sin90°)

B = 2.15/1.2

B = 1.79 T

Hence the average field strength is 1.79 T