Question

A square loop of side 1 m is placed in a perpendicular magnetic field.

Answer 1

A square loop of side 1 m is placed in a perpendicular magnetic field.Half of the area of the loop lies inside the magnetic field .A battery of emf 10 V and negligible internal resistance is connected in the loop.The magnetic field changes with time according to the relation B=(0.01-2t) T.Find the resultant emf of the circuit.

Answer 2

Answer:

9v

Explanation:

dB/dt= -2T

emf= -dphi/dt

= -d(BA)/dt

= -AdB/dt = -1/2(l^2) × (-2) = 1V

Since magnetic field (×) decreasing so according to Lenz’s law direction of induced current in upper part of square will be clockwise i.e. from A

to C or in oth er words emf induces in a direction opposite to the main emf so resultant emf = 10 – 1 = 9V.