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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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Approximately 5.9\; {\rm m\cdot s^{-1}} (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

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Overall, the nodes and antinodes on this spring would be:

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The distance between two adjacent nodes is equal to one-half (that is, (1/2)) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, (1/4)) of the wavelength of the wave.

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\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}.

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\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}.

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