A linear conductor of length l, located perpendicular to a magnetic field, is acted upon by a force F if the current in the conductor is I. With what force will the magnetic field act on: a) a conductor of length l + L bent at an angle \varphi, if the plane of the bend is perpendicular to the magnetic field, and the current in the conductor is I_1 b) a conductor in the form of a semicircle of radius R, through which a current I_2 flows, if the plane of the semicircle is perpendicular to the magnetic field?

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#9.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.1.3.$ A linear conductor of length $l$, located perpendicular to a magnetic field, is acted upon by a force $F$ if the current in the conductor is $I$. With what force will the magnetic field act on: </p><p>

a) a conductor of length $l + L$ bent at an angle $\varphi$, if the plane of the bend is perpendicular to the magnetic field, and the current in the conductor is $I_1$

b) a conductor in the form of a semicircle of radius $R$, through which a current $I_2$ flows, if the plane of the semicircle is perpendicular to the magnetic field?

    <h3>Solution</h3>
    <p>
        a) From Ampere's law, we find the magnetic field induction $B$

$$F_1 = Il \cdot B \Rightarrow B = \frac{F}{Il}\quad(1)$$
in this case, the numbers acting on the part of the wire with currents $I_1$ and $I_2$, respectively
$$F_1 = I_1l \cdot B; \quad F_2 = I_1L\cdot B$$
We substitute $(1)$:
$$F_1= \frac{I_1}{I}F; \quad F_2 = \frac{I_1L}{Il}F$$
The force acting on a piece of wire bent at an angle $\varphi$ can be found from the cosine law
$$|\vec{F_p}| = |\vec{F_1} + \vec{F_2}| = \sqrt{{F_1}^2+{F_2}^2+2 F_1 F_2\cdot\cos\varphi}$$

$$\boxed{|\vec{F_p}| = F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\frac{L}{l}\cdot\cos\varphi}}$$
b) For a conductor made of a semicircle, we will use the same formula for the magnetic field induction $B$
$$B = \frac{F}{Il}$$
On a small part of a conductor of length $dl$, the Ampere force acts
$$dF = B I_2~dl = B I_2R~d\theta$$
To find the force $F$, we integrate the forces acting on each part of the semiring.
$$F = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta dF = BI_2R\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta~d\theta$$
After mathematical transformations, we obtain the force acting on the semiring
$$\boxed{F = 2BI_2R = 2F\frac{I_2R}{Il}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\cos\varphi\frac{L}{l}};\quad2F\frac{I_2R}{Il}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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