MHT CET Medical202622 April 2026Evening ShiftPhysicsElectromagnetic InductionActual
A conducting rod AB is in contact with metal rails AC and BD, which are 0.25 m apart in a uniform magnetic field of induction 0.4 T acting perpendicular to the plane of the figure. Ends C and D are connected through a 5 resistor. If the rod AB moves to right with a velocity of 5 m/s, then the magnitude and direction of the current through rod AB is
Options
- A0.1 A from A to B
- B0.1 A from B to A
- C1 A from A to B
- D1 A from B to A
Correct answer
B. 0.1 A from B to A
Step-by-step solution
The motional emf induced in the moving rod is given by: E = Bvl Substituting the given values ( B = 0.4 T , v = 5 m/s , l = 0.25 m ): E = 0.4 5 0.25 = 0.5 V The magnitude of the induced current is: I = E R I = 0.5 5 = 0.1 A The direction of the induced current can be determined using Fleming's Right-Hand Rule or the Lorentz force on positive charges ( F = q( v B ) ). With velocity v towards the right and magnetic field B into the plane of the paper, the force on positive charges is directed upwards. Therefore, the