A straight section of a solenoid lead of length 0.39 m carries current 2.45 A in a magnetic field of flux density 0.32 T. The wire makes an angle 56° with the field. Calculate the magnetic force.

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Accepted Answer

0.2535 N Only the conductor length perpendicular to the magnetic field contributes to the force, so its magnitude is F = BIL sin θ. Substitute the field, current, length and stated angle: F = 0.32 T × 2.45 A × 0.39 m × sin 56°. Evaluate the angular factor: sin 56° = 0.829. Therefore F = 0.2535 N. Since the wire is neither parallel nor perpendicular to the field, the result lies between zero and BIL, as expected.

Force on a solenoid lead

A straight section of a solenoid lead of length 0.39 m carries current 2.45 A in a magnetic field of flux density 0.32 T. The wire makes an angle 56° with the field. Calculate the magnetic force.

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