a) A wire frame in the form of a circle with current can rotate about the horizontal axis OO^\prime. The mass of a unit length of the wire is \rho, the current in the frame is I. The frame is in a magnetic field of induction B, directed along the gravitational field. Determine the angle of deviation of the plane of the circle from the vertical. b) A wire frame in the form of a circle has a wire jumper along the diameter, parallel to the horizontal axis OO^\prime, about which the frame can rotate. The mass of a unit length of the frame and the jumper is the same and is equal to \rho. The current entering the frame is equal to I. The frame is in a magnetic field of induction B, directed parallel to the gravitational field. At what angle from the vertical will the frame deviate?

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <h3 id="back-link"><a href="/#9.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.1.11.$ a) A wire frame in the form of a circle with current can rotate about the horizontal axis $OO^\prime$. The mass of a unit length of the wire is $\rho$, the current in the frame is $I$. The frame is in a magnetic field of induction B, directed along the gravitational field. Determine the angle of deviation of the plane of the circle from the vertical. </p><p>

b) A wire frame in the form of a circle has a wire jumper along the diameter, parallel to the horizontal axis $OO^\prime$, about which the frame can rotate. The mass of a unit length of the frame and the jumper is the same and is equal to $\rho$. The current entering the frame is equal to $I$. The frame is in a magnetic field of induction $B$, directed parallel to the gravitational field. At what angle from the vertical will the frame deviate?

    <h3>Solution</h3>
    <p>
        a) Using the expression for the moment of magnetic field obtained in <a href="/9.1.10">9.1.10</a>, $\vec{N} = [\vec{M} \times \vec{B}]$ we obtain

$$N = BIS\cdot \cos\alpha = \pi R^2BI \cdot \cos\alpha$$
in this case the moment of gravity:
$$M_{mg} = Rmg\cdot \cos\alpha = 2\pi R^2\rho g\cdot \sin\alpha$$
these moments are equal:
$$2\pi R^2\rho g \cdot \sin\alpha = \pi BI R^2 \cdot \cos\alpha$$
after transformations, we obtain
$$\boxed{\tan\alpha = \frac{BI}{2\rho g}}$$
b) Find the moment of gravity:
$$M_{mg} = mgR\cdot \sin\alpha = (2\pi R + 2R)\rho gR \cdot \sin\alpha$$

$$M_{mg} = 2R^2\rho g(1+\pi ) \cdot \sin\alpha$$
Now, we represent the initial circuit as two superimposed circuits with currents $I_1$ and $I_2$ such that $I = I_1 + I_2$, the circuit with $I_2$ is a semicircle with a partition, the circuit with $I_1$ is a ring without a partition, then from the parallel connection it follows that: $I_1\cdot \pi R = I_2 \cdot 2R$ then
$$I_1 = I_2\cdot \frac{2}{\pi}$$

$$I = I_1 + I_2 = I_2\left(1+\frac{2}{\pi}\right)$$

$$I_1 = I\cdot \frac{2}{\pi +2}; \quad$$
Then the moments for these contours are:
$$M_{k1} = BS_1\cdot \cos\alpha =\pi R^2I\cdot \frac{2}{\pi +2}B\cdot \cos\alpha$$

$$M_{k2} = BS_2\cdot \cos\alpha = \frac{1}{2}\pi R^2I\cdot \frac{\pi}{\pi +2}B\cdot \cos\alpha$$
Total moment of the contours:
$$M_k = \frac{1}{2}\pi R^2 BI \cdot \cos\alpha\cdot \frac{\pi + 4}{\pi +2}$$
Let's equate this moment and the moment of gravity:
$$\frac{1}{2}\pi R^2 BI \cdot \cos\alpha\cdot \frac{\pi + 4}{\pi +2} = 2R^2\rho g(1+\pi ) \cdot \sin\alpha$$
from here
$$\boxed{\tan\alpha = \frac{\pi (\pi +4)}{4(\pi +2)(\pi +1)} \cdot \frac{BI}{\rho g}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$\tan\alpha = \frac{\pi (\pi +4)}{4(\pi +2)(\pi +1)} \cdot \frac{BI}{\rho g}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>