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timofeeve [1]
2 years ago
6

A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a

magnetic field of 0. 35 t, what is its potential energy?
Physics
1 answer:
Digiron [165]2 years ago
7 0

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>
  • We have the expression for torque acting on a current loop in a uniform magnetic field as,

                         \tau=MBsin\theta

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

  • As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
  • Thus, the potential energy will be,

            \tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

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Answer:

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Explanation:

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using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

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b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

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