Question
A thin conducting rod of length 3 m is aligned
vertically (along the z-axis) and is moving in space with a velocity vector v = (i + 4j) m/s within a uniform magnetic field given by B = (3i+j) Wb/m2 What is the magnitude of the emf induced between the ends of the rod?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.
Select the number from the given options which will replace the question mark in the following series.
45, 53, ? , 126, 151, 367
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