A loop of wire with a weight of 0.55 N is oriented vertically and carries a current I = 2.25 A. A segment of the wire passes through a magnetic field directed into the plane of the page as shwon. The net force on the wire is measured using a balance and found to be zero. What is the magnitude of the magnetic field?

Incomplete question.

Check the diagram given in attachment

Explanation:

Given that,

Weight w= 0.55N

Current I = 2.25A

Length = 0.2m

The net force is 0 N

Apply Newton second law

Fnet = ΣF

ΣF = 0

The weight is acting downward outside the plane

Then,

Magnetic Force — Weight = 0

F — W = 0

F = W

Magnetic force is given as

F = iLB

Therefore

iLB =W

B = W / iL

B = 0.55 / (2.25 × 0.2)

B = 1.22 T

The magnitude of the magnetic field is 1.22 T

jupiter

explanation:

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c. newton's third law of motion