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A horizontal wire of length 0.53 m. carrying a current of 7.5 A. is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of 19°, it experiences a magnetic field of 4.4x10^-3N. Determine the magnitude of the external magnetic field.

Respuesta :

Answer:

3.4 mT

Explanation:

L = 0.53 m

i = 7.5 A

Theta = 19 degree

F = 4.4 × 10^-3 N

Let B be the strength of magnetic field.

Force on a current carrying conductor placed in a magnetic field.

F = i × L × B × Sin theta

4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19

B = 3.4 × 10^-3 Tesla

B = 3.4 mT

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