Let's solve the problem step by step as requested: --- ## Step 1: Find the Magnetic Force (Part a) ### Explanation: A charge \( q \) at a distance \( r \) from the pivot rotates with velocity \( v = \omega_ r \) (since \( v = \omega r \) for circular motion). The magnetic force on the charge is given by: \[ \vec{F}_B = q \vec{v} \times \vec{B} \] Since \( \vec{v} \) is tangential and \( \vec{B} \) is into the page, the force points along the length of the bar. So, \[ F_B = q v B_ = q (\omega_ r) B_ \] **End of Step 1:** *The magnetic force on a charge at distance \( r \) from the pivot is:* \[ F_B = q \omega_ r B_ \] --- ## Step 2: Find the Potential Difference (Parts c & d) ### Explanation: At equilibrium, the electric force balances the magnetic force: \[ qE = q\omega_ r B_ \implies E = \omega_ r B_ \] The direction of \( E \) is along the bar, from the pivot outward. The potential difference is: \[ \Delta V = -\int_^L E \, dr = -\int_^L (\omega_ r B_) dr \] \[ = -\omega_ B_ \int_^L r \, dr = -\omega_ B_ \left[ \frac{r^2}{2} \right]_^L \] \[ = -\omega_ B_ \frac{L^2}{2} \] **End of Step 2:** *The potential difference between the ends of the bar is:* \[ \boxed{\Delta V = -\frac{1}{2} \omega_ B_ L^2} \] --- ### Final Answer: \[ \boxed{\Delta V = -\frac{1}{2} \omega_ B_ L^2} \] This is the potential difference between the ends of the bar when charge migration has ceased.

# Solution to Magnetic Field Problem We will solve the problem in two steps as outlined. ## Step 1: Magnetic Force on a Charge ### Explanation A charge \( q \) located at a distance \( r \) from the pivot of the rotating bar experiences a magnetic force due to the motion in the magnetic field \( B_u \). The velocity \( v \) of the charge is related to the angular velocity \( \omega \) and the radius \( r \) by the equation: \[ v = \omega r \] The magnetic force \( \vec{F}_B \) acting on the charge is given by: \[ \vec{F}_B = q \vec{v} \times \vec{B} \] Assuming the magnetic field \( B \) is directed perpendicularly to the plane of rotation, the magnitude of the magnetic force becomes: \[ F_B = q v B_u = q (\omega r) B_u \] **End of Step 1:** *The magnetic force acting on a charge at distance \( r \) from the pivot is given by:* \[ F_B = q \omega r B_u \] --- ## Step 2: Potential Difference Across the Bar ### Explanation When the bar rotates, charges within it will migrate until they reach an equilibrium state, where the electric force \( qE \) equals the magnetic force \( qF_B \): \[ qE = q\omega r B_u \implies E = \omega r B_u \] The electric field \( E \) is directed along the length of the bar from the pivot outward. To find the potential difference \( \Delta V \) across the length \( L \) of the bar, we integrate the electric field: \[ \Delta V = -\int_0^L E \, dr = -\int_0^L (\omega r B_u) \, dr \] Calculating the integral: \[ \Delta V = -\omega B_u \int_0^L r \, dr = -\omega B_u \left[ \frac{r^2}{2} \right]_0^L = -\omega B_u \frac{L^2}{2} \] **End of Step 2:** *The potential difference between the ends of the bar is:* \[ \Delta V = -\frac{1}{2} \omega B_u L^2 \] --- ### Final Answer The potential difference across the ends of the bar is: \[ \boxed{\Delta V = -\frac{1}{2} \omega B_u L^2} \]