A straight wire of length 3.5 m carries a current of 42 A and it is placed on a horizontal table in a horizontal uniform magnetic field.The wire makes an angle of 44 $\circ$ with the magnetic field lines.If the magnitude of the magnetic force on the wire is 9.0 N,what is the magnitude of the magnetic field?

Question

Quizplus · Accepted Answer

The magnetic force on a wire is given by the formula F = BILsinθ, where B is the magnetic field strength, I is the current, L is the length of the wire, and θ is the angle between the wire and the magnetic field lines.

In this case, we are given that F = 9.0 N, I = 42 A, L = 3.5 m, and θ = 44°.

Plugging these values in to the formula, we get:

9.0 N = B(42 A)(3.5 m)sin(44°)

Solving for B, we get:

B = 0.088 T

Therefore, the magnitude of the magnetic field is 0.088 T.

The correct answer is (C).

A Straight Wire of Length 3 ∘\circ∘ with the Magnetic Field Lines

A Straight Wire of Length 3 $\circ$ with the Magnetic Field Lines