A loop is pulled with a force F to the right to maintain a constant speed of 8.0 m/s. The loop has a length of 0.15 m, a width of 0.080 m, and a resistance of 200.0 $\varOmega$. At the instant shown, the loop is partially in and partially out of a uniform magnetic field that is directed into the paper. The magnitude of the field is 1.2 T. -Determine the magnitude of the force required to pull the loop. A)1.3 × 10^-⁴ N B)2.1 × 10^-⁴ N C)3.7 × 10^-⁴ N D)6.8 × 10^-⁴ N E)9.0 × 10^-⁴ N

Question

Quizplus · Accepted Answer

The force required to pull the loop is given by the equation:

F = BIL

where B is the magnitude of the magnetic field, I is the current in the loop, and L is the length of the loop in the direction perpendicular to the magnetic field.

Since the loop is moving with a constant velocity, the current in the loop must be equal to the emf divided by the resistance:

I = EMF/R

where EMF is the electromotive force or voltage induced in the loop by the magnetic field.

The emf is given by Faraday's law of induction:

EMF = BLv

where v is the speed of the loop perpendicular to the magnetic field.

Substituting these equations into the expression for the force:

F = B(IL) = B(EMF/R)L = (B^2Lv)/R

Plugging in the given values, we get:

F = (1.2 T)^2 (0.080 m)(8.0 m/s) / 200.0 Ω = 3.7 × 10^-4 N = 3.7 × 10^-1 mN ≈ 0.37 mN

Therefore, the best answer choice is C.

A Loop Is Pulled with a Force F to the Right