A thin conducting rod of length 3 m is aligned

Solution

The induced emf in a moving conductor in a magnetic field is given by the formula: ϵ=(v×B) ⋅ L where v is the velocity vector of the conductor, B is the magnetic field vector, and L is the vector representing the length and orientation of the conductor.     Given: ·        Length of the rod, l=3m. ·        The rod is aligned vertically along the z-axis, so the length vector is L=3km. ·        Velocity vector, v=i+4jm/s. ·        Magnetic field vector, B=3i+jWb/m First, calculate the cross-product v×B: Now, calculate the induced emf ϵ=(v×B) ⋅ L: ϵ=(−11k) ⋅ (3k) ϵ=(−11)×(3)(k ⋅ k) Since k ⋅ k=1, ϵ=−33V. The magnitude of the induced emf is ∣ ϵ ∣ = ∣ −33 ∣ =33V.