Let's solve the problems step by step, with explanations: --- ### **Part a) Expression for Magnetic Flux through the Rectangular Loop** #### **Step 1: Find the Magnetic Field at Distance \( r \) from the Wire** The magnetic field at a distance \( r \) from a long, straight wire carrying current \( I \) is given by: \[ B(r) = \frac{\mu_0 I}{2\pi r} \] where \( \mu_0 \) is the permeability of free space. #### **Step 2: Calculate the Magnetic Flux through the Loop** - The side of length \( a \) is parallel to the wire, and the loop is at distance \( d \) from the wire. - The magnetic field varies with distance from the wire, so we must integrate across the width \( b \) of the loop. The flux through a thin strip of width \( dx \) at distance \( x \) from the wire is: \[ d\Phi = B(x) \cdot (a \cdot dx) \] where \( x \) ranges from \( d \) to \( d+b \). So, total flux: \[ \Phi = \int_{d}^{d+b} \frac{\mu_0 I}{2\pi x} a \, dx \] \[ \Phi = \frac{\mu_0 I a}{2\pi} \int_{d}^{d+b} \frac{1}{x} dx \] \[ \Phi = \frac{\mu_0 I a}{2\pi} [\ln(x)]_{d}^{d+b} \] \[ \Phi = \frac{\mu_0 I a}{2\pi} \ln\left( \frac{d+b}{d} \right) \] **Final Expression for Magnetic Flux:** \[ \boxed{\Phi = \frac{\mu_0 I a}{2\pi} \ln\left( \frac{d+b}{d} \right)} \] --- ### **Part b) Induced emf in the Loop** Given: - \( \frac{dI}{dt} = 2\,A/s \) - \( a = 2\,cm = 0.02\,m \) - \( b = 5\,cm = 0.05\,m \) - \( d = 1\,cm = 0.01\,m \) - \( \mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A \) #### **Step 1: Find the emf using Faraday's Law** Faraday's Law: \[ \text{emf} = -\frac{d\Phi}{dt} \] From part (a), only \( I \) is changing with time: \[ \frac{d\Phi}{dt} = \frac{\mu_0 a}{2\pi} \ln\left( \frac{d+b}{d} \right) \frac{dI}{dt} \] So, \[ \text{emf} = - \frac{\mu_0 a}{2\pi} \ln\left( \frac{d+b}{d} \right) \frac{dI}{dt} \] #### **Step 2: Plug in the Values** \[ \text{emf} = - \frac{4\pi \times 10^{-7} \cdot 0.02}{2\pi} \ln\left( \frac{0.01 + 0.05}{0.01} \right) (2) \] \[ = - \frac{4\pi \times 10^{-7} \cdot 0.02}{2\pi} \ln(6) \cdot 2 \] \[ = - (2 \times 10^{-7} \cdot 0.02) \ln(6) \cdot 2 \] \[ = - (4 \times 10^{-9}) \ln(6) \cdot 2 \] \[ = - (8 \times 10^{-9}) \ln(6) \] \[ \ln(6) \approx 1.792 \] \[ = - (8 \times 10^{-9}) \times 1.792 \] \[ = -1.4336 \times 10^{-8}\, V \] **Final Answer:** \[ \boxed{\text{emf} \approx -1.4 \times 10^{-8} \, \text{V}} \] - The negative sign shows the direction of emf (Lenz's Law). --- **Summary:** - **(a) Magnetic flux:** \(\boxed{\frac{\mu_0 I a}{2\pi} \ln\left( \frac{d+b}{d} \right)}\) - **(b) Induced emf:** \(\boxed{-1.4 \times 10^{-8} \, \text{V}}\) for the given values.