A particle (q = –4.0 µC, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that i

Question

A particle (q = –4.0 µC, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s2. What is the magnitude of the magnetic field?

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Lệ Thu 4 years 2021-07-20T22:43:59+00:00 1 Answers 586 views 0

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  1. RobertKer
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    2021-07-20T22:45:53+00:00
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    Answer:

    The magnitude of the magnetic field is 4.73 T.

    Explanation:

    Given that,

    Charge of the particle, q=-4\ \mu C=-4\times 10^{-6}\ C

    It is moving with a velocity of 2 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s². We need to find the magnitude of magnetic field.

    When it enters in magnetic field,

    qvB\sin \theta=ma

    B is magnetic field

    B=\dfrac{ma}{qv\sin \theta}\\\\B=\dfrac{5\times 10^{-3}\times 5.8}{4\times 10^{-6}\times 2\times 10^3\sin (50)}\\\\B=4.73\ T

    So, the magnitude of the magnetic field is 4.73 T.

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