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Acharged particle moving through a magnetic field at right angles to the field with a speed of 36.2 m/s experiences a magnetic force of 7.38x104 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 12.1 m/s at an angle of 17° relative to the magnetic field. Express your answer in microNewtons.

Respuesta :

Answer:

7212.3 N

Explanation:

F = 7.38 x 10^4 N, v = 36.2 m/s

Let b be the strength of magnetic field and charge on the particle is q.

F = q v B Sin theta

Here theta = 90 degree

7.38 x 10^4 = q x 36.2 x B x Sin 90

q B = 2038.7 .....(1)

Now, theta = 17 degree, v = 12.1 m/s

F = q v B Sin theta

F = 2038.7 x 12.1 x Sin 17     ( q v = 2038.7 from equation (1)

F = 7212.3 N

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