Question

A straight segment of wire 35.0 cm long carrying a current of 2.60 A is in a uniform magnetic field. The segment makes an angle of 53∘ with the direction of the magnetic field. 1) If the force on the segment is 0.200 N, what is the magnitude of the magnetic field? (Express your answer to three significant figures.)

Answer 1

Answer:

Magnetic field, B = 0.275 T

Explanation:

Given that,

Length of the wire, L = 35 cm = 0.35 m

Current carried in the wire, I = 2.6 A

The segment makes an angle of 53∘ with the direction of the magnetic field, $\theta=53^{\circ}$

Magnetic force, F = 0.2 N

To find,

The magnitude of the magnetic field.

Solution,

The magnetic force acting on the wire is given by :

$F=ILBsin\theta$

$\theta$ is the angle between the length of wire and the magnetic field.

$0.2=2.6\times 0.35\times B\times sin(53)$

B = 0.275 T

Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.