# Problem 9 Solution Let's solve it step by step as requested: --- ## **Step 1: (A) Magnetic Flux through the Sheet** **Given:** - Current as a function of time: \[ I_{sp}(t) = I_ \frac{t}{\tau} \] - The sheet extends from the center to the edge (radius \( r \)), length \( W \), width \( r \). - We want the magnetic flux **through the sheet** as a function of time. **Explanation:** - The magnetic field inside a long straight wire (Ampère's Law): \[ B(r, t) = \frac{\mu_ I_{sp}(t)}{2\pi r} \] - The magnetic flux through the sheet is: \[ \Phi(t) = \int_^r B(r', t) \cdot W \, dr' \] - Substitute \( B(r', t) \) and integrate from to \( r \): \[ \Phi(t) = \int_^r \frac{\mu_ I_{sp}(t)}{2\pi r'} W \, dr' = \frac{\mu_ W I_{sp}(t)}{2\pi} \int_^r \frac{1}{r'} dr' = \frac{\mu_ W I_{sp}(t)}{2\pi} [\ln r' ]_^r \] - The lower limit diverges, but physically the wire has a small radius \( a \), so: \[ \Phi(t) = \frac{\mu_ W I_{sp}(t)}{2\pi} \ln\left(\frac{r}{a}\right) \] - Substitute \( I_{sp}(t) \): \[ \boxed{ \Phi(t) = \frac{\mu_ W I_}{2\pi \tau} t \ln\left(\frac{r}{a}\right) } \] **End of Step 1 Explanation:** We derived the time-dependent magnetic flux through the sheet, which increases linearly with time. --- ## **Step 2: (B) Induced Current in the Loop** **Given:** - Resistance of the loop: \( R_ = 1.00\,\Omega \) - Use Faraday’s Law: \[ \mathcal{E} = -\frac{d\Phi}{dt} \] - Induced current: \[ I_{\text{ind}} = \frac{\mathcal{E}}{R_} \] **Explanation:** - Differentiate \( \Phi(t) \) with respect to time: \[ \frac{d\Phi}{dt} = \frac{\mu_ W I_}{2\pi \tau} \ln\left(\frac{r}{a}\right) \] - So, the induced emf is: \[ \mathcal{E} = -\frac{\mu_ W I_}{2\pi \tau} \ln\left(\frac{r}{a}\right) \] - The induced current: \[ I_{\text{ind}} = \frac{\mathcal{E}}{R_} = -\frac{\mu_ W I_}{2\pi \tau R_} \ln\left(\frac{r}{a}\right) \] **End of Step 2 Explanation:** We found the induced current in the wire loop in terms of the given parameters. --- ## **Step 3: (C) Could this be an issue for a spacecraft?** **Explanation:** If the induced current is large enough, it could interfere with sensitive electronics or even damage equipment, especially if the loop encloses a large changing magnetic flux. In space, where shielding is difficult, this could be a significant problem for spacecraft operation. --- ## **Final Answers** ### **(A) Magnetic Flux:** \[ \boxed{ \Phi(t) = \frac{\mu_ W I_}{2\pi \tau} t \ln\left(\frac{r}{a}\right) } \] where \( a \) is the inner radius (wire thickness). --- ### **(B) Induced Current:** \[ \boxed{ I_{\text{ind}} = -\frac{\mu_ W I_}{2\pi \tau R_} \ln\left(\frac{r}{a}\right) } \] --- ### **(C) Spacecraft Issue:** \[ \boxed{ \text{Yes, this could be an issue for spacecraft, as induced currents can disrupt or damage sensitive electronics.} } \]

# Problem 9 Solution Let's solve it step by step as requested: --- ## **Step 1: (A) Magnetic Flux through the Sheet** **Given:** - Current as a function of time: \[ I_{sp}(t) = I_0 \left(1 - \frac{t}{\tau}\right) \] - The sheet extends from the center to the edge (radius \( r \)), length \( W \), and width \( r \). **Explanation:** - The magnetic field inside a long straight wire (using Ampère's Law): \[ B(r, t) = \frac{\mu_0 I_{sp}(t)}{2\pi r} \] - The magnetic flux through the sheet is given by: \[ \Phi(t) = \int_0^r B(r', t) \cdot W \, dr' \] - Substitute \( B(r', t) \) and integrate from \( 0 \) to \( r \): \[ \Phi(t) = \int_0^r \frac{\mu_0 I_{sp}(t)}{2\pi r'} W \, dr' = \frac{\mu_0 W I_{sp}(t)}{2\pi} \int_0^r \frac{1}{r'} dr' \] - This integral evaluates to: \[ \Phi(t) = \frac{\mu_0 W I_{sp}(t)}{2\pi} \ln\left(\frac{r}{a}\right) \] - Substitute \( I_{sp}(t) \): \[ \Phi(t) = \frac{\mu_0 W I_0 \left(1 - \frac{t}{\tau}\right)}{2\pi} \ln\left(\frac{r}{a}\right) \] - Thus, the expression for magnetic flux through the sheet as a function of time is: \[ \boxed{ \Phi(t) = \frac{\mu_0 W I_0}{2\pi} \left(1 - \frac{t}{\tau}\right) \ln\left(\frac{r}{a}\right) } \] **End of Step 1 Explanation:** We derived the time-dependent magnetic flux through the sheet, which decreases linearly over time due to the decreasing current. --- ## **Step 2: (B) Induced Current in the Loop** **Given:** - Resistance of the loop: \( R = 1.00 \, \Omega \) - Use Faraday’s Law: \[ \mathcal{E} = -\frac{d\Phi}{dt} \] - Induced current: \[ I_{\text{ind}} = \frac{\mathcal{E}}{R} \] **Explanation:** - Differentiate \( \Phi(t) \) with respect to time: \[ \frac{d\Phi}{dt} = \frac{\mu_0 W I_0}{2\pi} \left(-\frac{1}{\tau}\right) \ln\left(\frac{r}{a}\right) \] - The induced emf is then: \[ \mathcal{E} = -\frac{d\Phi}{dt} = \frac{\mu_0 W I_0}{2\pi \tau} \ln\left(\frac{r}{a}\right) \] - The induced current is given by: \[ I_{\text{ind}} = \frac{\mathcal{E}}{R} = \frac{\mu_0 W I_0}{2\pi \tau R} \ln\left(\frac{r}{a}\right) \] **End of Step 2 Explanation:** We found the induced current in the wire loop in terms of the given parameters. --- ## **Step 3: (C) Could this be an issue for a spacecraft?** **Explanation:** If the induced current is large enough, it could interfere with sensitive electronics or even damage equipment. In space, where shielding from electromagnetic interference is challenging, this could pose a significant risk for spacecraft operational integrity. --- ## **Final Answers** ### **(A) Magnetic Flux:** \[ \boxed{ \Phi(t) = \frac{\mu_0 W I_0}{2\pi} \left(1 - \frac{t}{\tau}\right) \ln\left(\frac{r}{a}\right) } \] --- ### **(B) Induced Current:** \[ \boxed{ I_{\text{ind}} = \frac{\mu_0 W I_0}{2\pi \tau R} \ln\left(\frac{r}{a}\right) } \] --- ### **(C) Spacecraft Issue:** \[ \boxed{ \text{Yes, this could be an issue for spacecraft, as induced currents can disrupt or damage sensitive electronics.} } \]