Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision		 Previous revision
electrical_engineering_2:the_time-dependent_magnetic_field [2024/05/07 03:52] – [Rod in Circuit] mexleadminelectrical_engineering_2:the_time-dependent_magnetic_field [2025/05/06 11:22] (current) – [Bearbeiten - Panel] mexleadmin
Line 48: Line 48:
 Therefore, the induced potential difference generated by a conductor or coil moving in a magnetic field is Therefore, the induced potential difference generated by a conductor or coil moving in a magnetic field is
  
-\begin{align*} \boxed{ U_{\rm ind} = -{{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} = -{{\rm d}\over{{\rm d}t}}\iint_A \vec{B} \cdot {\rm d} \vec{A} } \end{align*}+\begin{align*} \boxed{ u_{\rm ind} = -{{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} = -{{\rm d}\over{{\rm d}t}}\iint_A \vec{B} \cdot {\rm d} \vec{A} } \end{align*}
  
 The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter. The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter.
Line 91: Line 91:
 \begin{align*} \Phi_{\rm m} = B \cdot A \end{align*} \begin{align*} \Phi_{\rm m} = B \cdot A \end{align*}
  
-We can calculate the magnitude of the potential difference $|U_{\rm ind}|$ from Faraday’s law:+We can calculate the magnitude of the potential difference $|u_{\rm ind}|$ from Faraday’s law:
  
 \begin{align*}  \begin{align*} 
-|U_{\rm ind}| &= |-        {{\rm d}\over{{\rm d}t}}(N \cdot \Phi_{\rm m})| \\ +|u_{\rm ind}| &= |-        {{\rm d}\over{{\rm d}t}}(N \cdot \Phi_{\rm m})| \\ 
               &= |-N \cdot {{\rm d}\over{{\rm d}t}}(B \cdot A)           | \\                &= |-N \cdot {{\rm d}\over{{\rm d}t}}(B \cdot A)           | \\ 
               &= |-N \cdot l^2 \cdot {{{\rm d}B}\over{{\rm d}t}}| \\                &= |-N \cdot l^2 \cdot {{{\rm d}B}\over{{\rm d}t}}| \\ 
Line 104: Line 104:
  
 \begin{align*}  \begin{align*} 
-|I| &= {{ |U_{\rm ind}|}\over{R}} \\ +|I| &= {{ |u_{\rm ind}|}\over{R}} \\ 
     &= {{0.50 ~\rm V}\over{5.0 ~\Omega}} = 0.10 ~\rm A \\      &= {{0.50 ~\rm V}\over{5.0 ~\Omega}} = 0.10 ~\rm A \\ 
 \end{align*}  \end{align*} 
Line 199: Line 199:
  
 \begin{align*}  \begin{align*} 
-U_{\rm ind} &=   \int_l \vec{E}_{\rm ind}        \cdot {\rm d} \vec{s} \\ +u_{\rm ind} &=   \int_l \vec{E}_{\rm ind}        \cdot {\rm d} \vec{s} \\ 
             &= - \int^0_1 \vec{v} \times \vec{B} \cdot {\rm d} \vec{s} \\              &= - \int^0_1 \vec{v} \times \vec{B} \cdot {\rm d} \vec{s} \\ 
 \end{align*} \end{align*}
Line 205: Line 205:
 For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to:  For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to: 
 \begin{align*}  \begin{align*} 
-U_{\rm ind} &= - v \cdot B \cdot l \\ +u_{\rm ind} &= - v \cdot B \cdot l \\ 
 \end{align*} \end{align*}
  
Line 216: Line 216:
 The velocity of the rod is $v={\rm d}x/{\rm d}t$. So the induced potential difference will get  The velocity of the rod is $v={\rm d}x/{\rm d}t$. So the induced potential difference will get 
 \begin{align*}  \begin{align*} 
-U_{\rm ind} &= - v \cdot B \cdot l \\ +u_{\rm ind} &= - v \cdot B \cdot l \\ 
             &= - {{{\rm d}x}\over{{\rm d}t}} \cdot B \cdot l \\              &= - {{{\rm d}x}\over{{\rm d}t}} \cdot B \cdot l \\ 
             &= - {{B \cdot l \cdot {\rm d}x}\over{{\rm d}t}} \\              &= - {{B \cdot l \cdot {\rm d}x}\over{{\rm d}t}} \\ 
Line 225: Line 225:
 This is an alternative way to deduce Faraday's Law. This is an alternative way to deduce Faraday's Law.
  
-The current $I_{\rm ind}$ induced in the given circuit is $U_{\rm ind}$ divided by the resistance $R$+The current $i_{\rm ind}$ induced in the given circuit is $u_{\rm ind}$ divided by the resistance $R$
  
 \begin{align*}  \begin{align*} 
-I_{\rm ind} = {{v \cdot B \cdot l }\over{R}} +i_{\rm ind} = {{v \cdot B \cdot l }\over{R}} 
 \end{align*} \end{align*}
  
Line 236: Line 236:
  
 \begin{align*}  \begin{align*} 
-U_{\rm ind} &= - {{\rm d}\over{{\rm dt}}} \cdot \Phi_{\rm m} \\ +u_{\rm ind} &= - {{\rm d}\over{{\rm dt}}} \cdot \Phi_{\rm m} \\ 
             &= - B \cdot l \cdot {{{\rm d}x}\over{{\rm d}t}} \\              &= - B \cdot l \cdot {{{\rm d}x}\over{{\rm d}t}} \\ 
             &= - B \cdot l \cdot v \\              &= - B \cdot l \cdot v \\ 
Line 251: Line 251:
 <button size="xs" type="link" collapse="Solution_4_3_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_1_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_1_1_Solution" collapsed="true">
  
-This is a great example of using the equation motional $U_{\rm ind} = - B \cdot l \cdot v$+This is a great example of using the equation motional $u_{\rm ind} = - B \cdot l \cdot v$
  
-Entering the given values into $U_{\rm ind} = - B \cdot l \cdot v$ gives+Entering the given values into $u_{\rm ind} = - B \cdot l \cdot v$ gives
  
 \begin{align*}  \begin{align*} 
-U_{ind} &= - {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} \\ +u_{ind} &= - {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} \\ 
         &= - B \cdot l \cdot v \\          &= - B \cdot l \cdot v \\ 
         &= - (5.00 \cdot 10^{-5}~\rm T)(20.0 \cdot 10^{3} ~\rm m)(7.80\cdot 10^{3} ~\rm m/s) \\          &= - (5.00 \cdot 10^{-5}~\rm T)(20.0 \cdot 10^{3} ~\rm m)(7.80\cdot 10^{3} ~\rm m/s) \\ 
Line 264: Line 264:
  
 <button size="xs" type="link" collapse="Solution_4_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_1_1_Result" collapsed="true">  <button size="xs" type="link" collapse="Solution_4_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_1_1_Result" collapsed="true"> 
-\begin{align*} U_{\rm ind} &= - 7.80 \cdot 10^3 ~\rm V \\ \end{align*}+\begin{align*} u_{\rm ind} &= - 7.80 \cdot 10^3 ~\rm V \\ \end{align*}
  
 </collapse> </collapse>
Line 296: Line 296:
  
 \begin{align*}  \begin{align*} 
-U_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\ +u_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\ 
             &= B\cdot {{r^2\omega}\over{2}}              &= B\cdot {{r^2\omega}\over{2}} 
 \end{align*} \end{align*}
Line 303: Line 303:
  
 \begin{align*}  \begin{align*} 
-I_{\rm ind} &= {{|U_{\rm ind}|}\over{R}} \\ +i_{\rm ind} &= {{|u_{\rm ind}|}\over{R}} \\ 
             &= B\cdot {{r^2\omega}\over{2R}}              &= B\cdot {{r^2\omega}\over{2R}} 
 \end{align*} \end{align*}
Line 316: Line 316:
 The coil is rotated about the $z$-axis through its center at a constant angular velocity $\omega$. The coil is rotated about the $z$-axis through its center at a constant angular velocity $\omega$.
  
-Obtain an expression for the induced potential difference $U_{\rm ind}$ in the coil.+Obtain an expression for the induced potential difference $u_{\rm ind}$ in the coil.
  
 <WRAP> <imgcaption ImgNr12 | A rectangular coil rotating in a uniform magnetic field> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating2}} </WRAP> <WRAP> <imgcaption ImgNr12 | A rectangular coil rotating in a uniform magnetic field> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating2}} </WRAP>
Line 322: Line 322:
 <button size="xs" type="link" collapse="Solution_4_3_3_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_3_1_Strategy" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_3_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_3_1_Strategy" collapsed="true">
  
-According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $\cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration simplify quick. The induced potential difference is written out using Faraday’s law. </collapse>+According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $\cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration simpler. The induced potential difference is written out using Faraday’s law. </collapse>
  
 <button size="xs" type="link" collapse="Solution_4_3_3_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_3_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_3_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_3_1_Solution" collapsed="true">
Line 755: Line 755:
 \begin{align*}  \begin{align*} 
 L_4 &= \mu_0 \mu_{\rm r,4} \cdot N^2 \cdot {{A }\over {l}} \\ L_4 &= \mu_0 \mu_{\rm r,4} \cdot N^2 \cdot {{A }\over {l}} \\
-    &= \mu_0 \codt 1000 \cdot N^2 \cdot {{A }\over {l}} \\+    &= \mu_0 \cdot 1000 \cdot N^2 \cdot {{A }\over {l}} \\
     &= 1000 \cdot L_4 \\     &= 1000 \cdot L_4 \\
 \end{align*} \end{align*}