# Magnetic Flux and Induced Current Near a Long Wire Let's solve the problem step by step. ## **Given Data** - **Length of loop:** \( a = 3.4 \) cm = **.034 m** - **Width of loop:** \( b = 1. \) cm = **.010 m** - **Resistance of loop:** \( R = .41 \) mΩ = **.00041 Ω** (assuming "mΩ" means milliohms) - **Current in wire:** \( i = 5.3 \) A - **Distance from wire to center of loop:** \( r = 1.4 \) cm = **.014 m** - **Speed of loop:** \( v = 1.6 \) mm/s = **.0016 m/s** --- ## **(a) Magnetic Flux Through the Loop** ### **Step 1: Magnetic Field from Long Wire** The magnetic field at a distance \( x \) from a long straight wire: \[ B(x) = \frac{\mu_ i}{2\pi x} \] where \( \mu_ = 4\pi \times 10^{-7} \) T·m/A. ### **Step 2: Magnetic Flux Calculation** The magnetic flux through the loop: \[ \Phi = \int_{x_1}^{x_2} B(x) \cdot a \, dx \] - \( x_1 = r - \frac{b}{2} \) - \( x_2 = r + \frac{b}{2} \) So, \[ \Phi = a \int_{x_1}^{x_2} \frac{\mu_ i}{2\pi x} dx \] \[ \Phi = a \cdot \frac{\mu_ i}{2\pi} \left[ \ln x \right]_{x_1}^{x_2} \] \[ \Phi = a \cdot \frac{\mu_ i}{2\pi} \ln\left(\frac{x_2}{x_1}\right) \] #### **Plug in the numbers:** - \( x_1 = .014 - .005 = .009 \) m - \( x_2 = .014 + .005 = .019 \) m - \( a = .034 \) m - \( i = 5.3 \) A \[ \Phi = .034 \cdot \frac{4\pi \times 10^{-7} \times 5.3}{2\pi} \ln\left( \frac{.019}{.009} \right) \] \[ \frac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7} \] \[ \Phi = .034 \cdot (2 \times 10^{-7} \times 5.3) \cdot \ln\left( \frac{.019}{.009} \right) \] \[ 2 \times 10^{-7} \times 5.3 = 1.06 \times 10^{-6} \] \[ \ln\left( \frac{.019}{.009} \right) = \ln(2.111...) \approx .747 \] \[ \Phi = .034 \cdot 1.06 \times 10^{-6} \cdot .747 \] First, \( 1.06 \times 10^{-6} \cdot .747 \approx 7.928 \times 10^{-7} \) Then, \( .034 \cdot 7.928 \times 10^{-7} \approx 2.695 \times 10^{-8} \) Wb ### **Final Answer for (a):** \[ \boxed{ \Phi = 2.7 \times 10^{-8}~\text{Wb} } \] --- ## **(b) Induced Current in the Loop** ### **Step 1: Induced EMF (Faraday's Law)** \[ \mathcal{E} = \left| \frac{d\Phi}{dt} \right| \] Since the loop moves away at speed \( v \), \( r \) increases with time: \( r = r_ + vt \). \[ \mathcal{E} = \left| \frac{d\Phi}{dr} \cdot \frac{dr}{dt} \right| = v \left| \frac{d\Phi}{dr} \right| \] ### **Step 2: Compute \( d\Phi/dr \)** Recall: \[ \Phi(r) = a \cdot \frac{\mu_ i}{2\pi} \ln \left( \frac{r + b/2}{r - b/2} \right) \] Take derivative with respect to \( r \): \[ \frac{d\Phi}{dr} = a \cdot \frac{\mu_ i}{2\pi} \cdot \frac{d}{dr} \ln \left( \frac{r + b/2}{r - b/2} \right) \] \[ \frac{d}{dr} \ln\left( \frac{r + b/2}{r - b/2} \right) = \frac{1}{r + b/2} + \frac{1}{r - b/2} \] So, \[ \frac{d\Phi}{dr} = a \cdot \frac{\mu_ i}{2\pi} \left( \frac{1}{r + b/2} + \frac{1}{r - b/2} \right) \] ### **Step 3: Plug in the Numbers** - \( r = .014 \) m, \( b = .010 \) m, so \( b/2 = .005 \) m - \( r + b/2 = .019 \) m - \( r - b/2 = .009 \) m \[ \frac{1}{.019} + \frac{1}{.009} \approx 52.63 + 111.11 = 163.74 \] \[ a \cdot \frac{\mu_ i}{2\pi} = .034 \cdot 1.06 \times 10^{-6} = 3.604 \times 10^{-8} \] \[ \frac{d\Phi}{dr} = 3.604 \times 10^{-8} \cdot 163.74 \approx 5.90 \times 10^{-6}~\text{Wb/m} \] ### **Step 4: Induced EMF** \[ \mathcal{E} = v \left| \frac{d\Phi}{dr} \right| = .0016 \cdot 5.90 \times 10^{-6} \approx 9.44 \times 10^{-9}~\text{V} \] ### **Step 5: Induced Current** \[ I = \frac{\mathcal{E}}{R} = \frac{9.44 \times 10^{-9}}{.00041} \approx 2.3 \times 10^{-5}~\text{A} \] --- ## **Final Answers** ### **(a) Magnetic Flux Through the Loop** \[ \boxed{2.7 \times 10^{-8}~\text{Wb}} \] ### **(b) Induced Current in the Loop** \[ \boxed{2.3 \times 10^{-5}~\text{A}} \] --- ### **Diagram (Description)** *No image provided, but imagine a long straight wire with a small rectangular loop of wire positioned parallel to it, with the loop moving away from the wire.* --- #### **Formula Summary** - **Magnetic Field from wire:** \( B(x) = \frac{\mu_ i}{2\pi x} \) - **Flux:** \( \Phi = a \cdot \frac{\mu_ i}{2\pi} \ln\left( \frac{r+b/2}{r-b/2} \right) \) - **Induced EMF:** \( \mathcal{E} = v \left| \frac{d\Phi}{dr} \right| \) - **Induced Current:** \( I = \frac{\mathcal{E}}{R} \)