Translated 9.1.1-9.1.5
en/9.1.3.md
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| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
| + | <meta property="og:title" content="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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| + | <meta property="og:description" content="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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| + | <title>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?</title> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
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| + | </header> | ||
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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 $I_1$ </p><p> | ||
| + | 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? | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/9/9.1.3/statement.png" | ||
| + | loading="lazy" width="230" /> | ||
| + | <figcaption> | ||
| + | For problem $9.1.4$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <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> | ||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
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| <a href="../../" style="text-decoration: none;"> | |||
| <div id="logo"> | |||
| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
| </div> | |||
| </a> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <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 $I_1$ </p><p> | |||
| 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? | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.3/statement.png" | |||
| loading="lazy" width="230" /> | |||
| <figcaption> | |||
| For problem $9.1.4$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <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> | |||
| <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> | |||
| </body> | |||
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