A straight wire of length 5.0 m carries a current I and it is placed on a horizontal table in a horizontal uniform magnetic field of magnitude 0.016 T.The wire makes an angle of 25 $\circ$ with the magnetic field lines.If the magnitude of the magnetic force on the wire is 0.62 N,what is the current I?

Question

Quizplus · Accepted Answer

The formula for magnetic force on a current-carrying wire in a magnetic field is given by F = I * L * B * sin(theta), where I is the current, L is the length of the wire, B is the magnetic field strength, and theta is the angle between the wire and the magnetic field lines.

In this problem, we are given L = 5.0 m, B = 0.016 T, theta = 25 degrees, and F = 0.62 N. We need to solve for I.

Plugging in the given values, we get:

0.62 N = I * 5.0 m * 0.016 T * sin(25 degrees)

Solving for I, we get:

I = 18 A

Therefore, the answer is B.

A Straight Wire of Length 5 ∘\circ∘ with the Magnetic Field Lines

A Straight Wire of Length 5 $\circ$ with the Magnetic Field Lines