Translated 9.2.1-9.2.9
en/9.2.8.md
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| + | <meta name="description" content="a) Using the formula given in problem , determine the magnetic induction created by a charge q moving with velocity v at a distance r from this charge. The radius vector \vec{r} forms an angle \alpha with the velocity v. b) Determine the magnetic induction of a straight wire of length l carrying a current I at a distance r from the wire if l \ll r. The radius vector \vec{r} forms an angle \alpha with the wire."> | ||
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| + | <meta property="og:title" content="a) Using the formula given in problem , determine the magnetic induction created by a charge q moving with velocity v at a distance r from this charge. The radius vector \vec{r} forms an angle \alpha with the velocity v. b) Determine the magnetic induction of a straight wire of length l carrying a current I at a distance r from the wire if l \ll r. The radius vector \vec{r} forms an angle \alpha with the wire."> | ||
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| + | <title>a) Using the formula given in problem , determine the magnetic induction created by a charge q moving with velocity v at a distance r from this charge. The radius vector \vec{r} forms an angle \alpha with the velocity v. b) Determine the magnetic induction of a straight wire of length l carrying a current I at a distance r from the wire if l \ll r. The radius vector \vec{r} forms an angle \alpha with the wire.</title> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | Solutions of Savchenko Problems in Physics <br> | ||
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| + | <h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $9.2.8.$ a) Using the formula given in problem <a href="../9.2.1" target="_blank">9.2.1</a>, determine the magnetic induction created by a charge $q$ moving with velocity $v$ at a distance $r$ from this charge. The radius vector $\vec{r}$ forms an angle $\alpha$ with the velocity $v$. </p><p> | ||
| + | b) Determine the magnetic induction of a straight wire of length $l$ carrying a current $I$ at a distance $r$ from the wire if $l \ll r$. The radius vector $\vec{r}$ forms an angle $\alpha$ with the wire. | ||
| + | </p> | ||
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| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/9/9.2.8/statement.png" | ||
| + | loading="lazy" width="400" /> | ||
| + | <figcaption> | ||
| + | For problem $9.2.8$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | a) Let's write down the formula given in the problem <a href="../9.2.1" target="_blank">9.2.1</a>, | ||
| + | $$ \vec{B} = \mu_0 \varepsilon_0 [\vec{v} \times \vec{E}] \quad(1) $$ | ||
| + | We will find the current flow rate from <a href="../9.2.3" target="_blank">9.2.3</a> | ||
| + | $$ v = \frac{I}{\lambda}\quad(2) $$ | ||
| + | Considering that the charge $q = l\lambda$, the electric field strength | ||
| + | $$ E = \frac{l\lambda}{4\pi\varepsilon_0r^2}\quad(3) $$ | ||
| + | We substitute $(2)$ and $(3)$ into $(1)$, and we get the expression | ||
| + | $$ \vec{B} = \mu_0 \varepsilon_0\left[\vec{\left(\frac{I}{\lambda}\right)} \times \frac{l\lambda}{4\pi\varepsilon_0} \cdot \frac{\vec{r}}{r^3}\right] \quad(4) $$ | ||
| + | |||
| + | $$ \vec{B} = \frac{\mu_0 q}{4\pi r^3}[\vec{v} \times \vec{r}] $$ | ||
| + | The modulus of this vector product is | ||
| + | $$ B = \frac{\mu_0 q}{4\pi r^3}vr\cdot \sin\alpha \Rightarrow \boxed{B = \frac{\mu_0 qv}{4\pi r^2}\cdot \sin\alpha} $$ | ||
| + | </p><p> | ||
| + | b) Since $l \ll r$ then the approximation of the thread as a point charge is valid | ||
| + | $$ B = \mu_0 \varepsilon_0 Ev\cdot \sin\alpha $$ | ||
| + | We substitute the expressions $v$ and $E$ from $(2)$ and $(3)$, respectively | ||
| + | $$ B = \mu_0 \varepsilon_0 \frac{l\lambda}{4\pi\varepsilon r^2} \cdot \frac{I}{\lambda}\sin\alpha \Rightarrow \boxed{B=\frac{\mu_0Il}{4\pi r^2}\sin\alpha} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)$$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
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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="../../#9.2">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $9.2.8.$ a) Using the formula given in problem <a href="../9.2.1" target="_blank">9.2.1</a>, determine the magnetic induction created by a charge $q$ moving with velocity $v$ at a distance $r$ from this charge. The radius vector $\vec{r}$ forms an angle $\alpha$ with the velocity $v$. </p><p> | |||
| b) Determine the magnetic induction of a straight wire of length $l$ carrying a current $I$ at a distance $r$ from the wire if $l \ll r$. The radius vector $\vec{r}$ forms an angle $\alpha$ with the wire. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.2.8/statement.png" | |||
| loading="lazy" width="400" /> | |||
| <figcaption> | |||
| For problem $9.2.8$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| a) Let's write down the formula given in the problem <a href="../9.2.1" target="_blank">9.2.1</a>, | |||
| $$ \vec{B} = \mu_0 \varepsilon_0 [\vec{v} \times \vec{E}] \quad(1) $$ | |||
| We will find the current flow rate from <a href="../9.2.3" target="_blank">9.2.3</a> | |||
| $$ v = \frac{I}{\lambda}\quad(2) $$ | |||
| Considering that the charge $q = l\lambda$, the electric field strength | |||
| $$ E = \frac{l\lambda}{4\pi\varepsilon_0r^2}\quad(3) $$ | |||
| We substitute $(2)$ and $(3)$ into $(1)$, and we get the expression | |||
| $$ \vec{B} = \mu_0 \varepsilon_0\left[\vec{\left(\frac{I}{\lambda}\right)} \times \frac{l\lambda}{4\pi\varepsilon_0} \cdot \frac{\vec{r}}{r^3}\right] \quad(4) $$ | |||
| $$ \vec{B} = \frac{\mu_0 q}{4\pi r^3}[\vec{v} \times \vec{r}] $$ | |||
| The modulus of this vector product is | |||
| $$ B = \frac{\mu_0 q}{4\pi r^3}vr\cdot \sin\alpha \Rightarrow \boxed{B = \frac{\mu_0 qv}{4\pi r^2}\cdot \sin\alpha} $$ | |||
| </p><p> | |||
| b) Since $l \ll r$ then the approximation of the thread as a point charge is valid | |||
| $$ B = \mu_0 \varepsilon_0 Ev\cdot \sin\alpha $$ | |||
| We substitute the expressions $v$ and $E$ from $(2)$ and $(3)$, respectively | |||
| $$ B = \mu_0 \varepsilon_0 \frac{l\lambda}{4\pi\varepsilon r^2} \cdot \frac{I}{\lambda}\sin\alpha \Rightarrow \boxed{B=\frac{\mu_0Il}{4\pi r^2}\sin\alpha} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)$$ | |||
| </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> | |||
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