A conducting rod PQ of length l = 2m is moving at a speed of 2 ms-1 making an angle of 30° with its length…..

Q. A conducting rod PQ of length l = 2m is moving at a speed of 2 ms-1 making an angle of 30° with its length. A uniform magnetic field B = 2 T exists in a direction perpendicular to the plane of motion. Then

Numerical

(a) VP – VQ = 8V

(b) VP – VQ = 4V

(c) VQ – VP = 8V

(d) VQ – VP = 4V

Ans: (b)

For motional emf B , l & v should be mutually perpendicular

$ \displaystyle V_P – V_Q = B (PQ)(v sin\theta)$

$ \displaystyle V_P – V_Q = 2 (2)(2 \times sin30)$

$ \displaystyle V_P – V_Q = 2 (2)(2 \times \frac{1}{2})$

$ \displaystyle V_P – V_Q = 4 V $

Leave a Comment