Let's solve the problem step by step: --- ## **Given Data** - **Current in wire (\(I\))**: \(20\, \text{A}\) (in \(-z\) direction) - **Velocity of rod (\(u\))**: \(5a_z\, \text{m/s}\) (parallel to wire, assume direction is along positive \(z\)) - **Length of rod (\(PQ\))**: \(l = 50\,\text{cm} = .5\,\text{m}\) - **Distance from wire to rod**: \(r_1 = 15\,\text{cm} = .15\,\text{m}\) - **Rod is parallel to the wire** --- ## **Step 1: Find the Magnetic Field at the Rod** The magnetic field at distance \(r\) from a long straight wire: \[ B = \frac{\mu_ I}{2\pi r} \] where: - \(\mu_ = 4\pi \times 10^{-7}\, \text{T}\cdot\text{m/A}\) - \(I = 20\, \text{A}\) Since the rod is parallel to the wire, the magnetic field at distance \(r\) from the wire points into/out of the page (use right-hand rule). --- ## **Step 2: Induced EMF in the Moving Rod** The emf induced in a rod moving perpendicularly to a magnetic field is: \[ \mathcal{E} = \int_P^Q (\vec{v} \times \vec{B}) \cdot d\vec{l} \] - \(\vec{v}\) is along \(z\) (let's take out of the page), - \(\vec{B}\) is tangential (from right-hand rule, at the rod's position, it is along \(y\)), - \(d\vec{l}\) is along \(x\) (from \(P\) to \(Q\)), perpendicular to both. Let’s express position along the rod as \(x\), where \(x\) goes from \(r_1 = .15\,\text{m}\) to \(r_2 = r_1 + l = .15 + .5 = .65\,\text{m}\). \[ B(x) = \frac{\mu_ I}{2\pi x} \] --- ## **Step 3: Calculate the Induced EMF** \[ \mathcal{E} = \int_{x_1}^{x_2} u B(x) dx \] \[ \mathcal{E} = u \int_{x_1}^{x_2} \frac{\mu_ I}{2\pi x} dx \] \[ \mathcal{E} = u \frac{\mu_ I}{2\pi} \int_{x_1}^{x_2} \frac{dx}{x} \] \[ \mathcal{E} = u \frac{\mu_ I}{2\pi} \left[ \ln(x) \right]_{x_1}^{x_2} \] \[ \mathcal{E} = u \frac{\mu_ I}{2\pi} \ln\left( \frac{x_2}{x_1} \right) \] Plug in the values: - \(u = 5\,\text{m/s}\) - \(\mu_ = 4\pi \times 10^{-7}\) - \(I = 20\,\text{A}\) - \(x_1 = .15\,\text{m}\) - \(x_2 = .65\,\text{m}\) --- ### **Plugging in Values** \[ \mathcal{E} = 5 \cdot \frac{4\pi \times 10^{-7} \cdot 20}{2\pi} \cdot \ln\left( \frac{.65}{.15} \right) \] \[ = 5 \cdot \frac{4 \cdot 20 \cdot 10^{-7}}{2} \cdot \ln\left( \frac{.65}{.15} \right) \] \[ = 5 \cdot \frac{80 \times 10^{-7}}{2} \cdot \ln\left( \frac{.65}{.15} \right) \] \[ = 5 \cdot 40 \times 10^{-7} \cdot \ln\left( \frac{.65}{.15} \right) \] \[ = 200 \times 10^{-7} \cdot \ln\left( \frac{.65}{.15} \right) \] Calculate \(\ln(.65/.15)\): \[ \frac{.65}{.15} = 4.333 \] \[ \ln(4.333) \approx 1.464 \] So, \[ \mathcal{E} = 200 \times 10^{-7} \cdot 1.464 \] \[ = 292.8 \times 10^{-7} \] \[ = 2.93 \times 10^{-5} \text{ V} \] --- ## **Step 4: Which End is at Lower Potential?** - The induced emf is such that it opposes the change (Lenz's law). - Use right-hand rule: For a positive charge, \(\vec{v} \times \vec{B}\) gives the direction of emf. - \(\vec{v}\) is along \(z\), \(\vec{B}\) is along \(\phi\) (tangent at the rod, which is into the page at the rod's location). - For this setup, potential at Q is **lower** than at P. --- ## **Final Answers** ### **Induced emf in the rod:** \[ \boxed{2.93 \times 10^{-5} \ \text{V}} \] ### **End at lower potential:** \[ \boxed{\text{Q is at lower potential than P}} \] If you need a detailed vector explanation for the direction, let me know!