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 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?

a) 0.18 T
b) 0.45 T
c) 1.08 T
d) 2.16 T

Answer 1

The average magnetic field strength is found using the formula F = BIL, where F is the force, I is the current, and L is the length of the wire.

With an input of 2.16 N for force, 30.0 A for current, and 0.04 m for length of wire, the magnetic field strength is calculated to be 0.18 T.

Step-by-step explanation:

To find the average field strength, we can use the formula for the magnetic force on a current-carrying conductor placed in a magnetic field: F = BILsin(θ), where F is the force in newtons, B is the magnetic field strength in teslas, I is the current in amperes, L is the length of the wire in the magnetic field in meters, and θ is the angle between the field and the current.

In this case, since the wire is perpendicular to the field, θ is 90 degrees and sin(θ) is 1, simplifying the formula to F = BIL. We're given F = 2.16 N, I = 30.0 A, and L = 4.00 cm = 0.04 m.

By rearranging the formula to solve for B, we have B = F/(IL). Plugging in the given values, B = 2.16 N / (30.0 A × 0.04 m) which gives us the magnetic field strength of 1.80 T, or 0.18 T in terms of the given answer choices.