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An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

Respuesta :

aabdulsamir

Answer: 0.297N

Explanation:

B = 0.550T

r = 2.50cm = 0.025m

I = 10.8A

L = 2r = 2* 0.025 = 0.050m

¤(theta) = 90°

Magnitude of the force (F) = BIL Sin ¤

F = 0.550 * 10.8 * 0.050 * sin 90

F = 0.550 * 10.8 * 0.050 * 1

F = 0.297N

onyebuchinnaji

The magnitude of magnetic force the field exert on the wire is 0.297 N.

The given parameters;

  • magnetic field strength, B = 0.55 T
  • radius of the cylinder, r = 2.5 cm = 0.025 m
  • current in the wire, I = 10.8 A

The magnitude of magnetic force the field exert on the wire is calculated as follows;

F = BILsin(θ)

where;

θ is the direction of the current = 90⁰

F = BI(2r)

F = 0.55 x 10.8 x 2 x 0.025

F = 0.297 N

Thus, the magnitude of magnetic force the field exert on the wire is 0.297 N.

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