A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic field (provi

Question

A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic field (provided by current-carrying coils) with a magnitude of 0.0800 T, which makes an angle of 35.0° with the orientation of the permanent magnet.

a. What is the magnitude of the torque (in N · m) on the permanent magnet?
b. What is the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils?

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Phúc Điền 7 months 2021-07-15T01:51:17+00:00 1 Answers 12 views 0

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    2021-07-15T01:52:43+00:00

    Answer:

    the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

    the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

    Explanation:

    The torque is given by :

    \bar {N} = \bar {m} * \bar {B}

    where ;

    m = 0.160 A.m²

    B = 0.0800 T

    θ = 35°

    So the magnitude of the torque N = mBsinθ

    N = (0.160)(0.0800)(sin 35°)

    N = 0.007341

    N = 7.34×10⁻³ Nm

    Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

    b) The potential energy \bar{U} = \bar{-m} * \bar{B}

    U = -mBcosθ

    U = (- 0.160)(0.0800)(cos 45)

    U = -0.010485

    U = -1.0485 ×10⁻² J

    Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

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