Question

A conducting rod AB of length l = 1 m is moving at a velocity v = 4 m/s making an angle 30^o with it’s length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motion. Then

• A. V_A – V_B = 8 V
• B. V_A – V_B = 4 V
• C. V_B – V_A = 8 V
• D. V_B – V_A = 4 V

Answer 1

Answer: (B)

V_A – V_B = 4 V

Answer 2

The emf induced across the rod AB is e = Bv_⊥l

Here v_⊥ = v sin 30⁰ = component of velocity perpendicular to length. Hence$e = Bvl\sin 30^{o} = (2)(4)(1)\left( \frac{1}{2} \right) = 4V$

The free electrons of the rod shift towards right due to the force $q(\overrightarrow{v} \times \overrightarrow{B}).$ Thus the left side of the rod is at higher potential. or V_A – V_B = 4V