A square wire loop with 2 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field (see Fig.) The loop contains a 20 V battery with negligible internal resistance. If the magnitude of the field varies with time according to B = (2 + 5t), with B in tesla and t in second.

What is the current through the battery ?

Options

A. `1/3 A`
B. `1/2 A`
C. `1 A`
D. `3/2 A`

Answer

C

Explanation

`a=2m`
`B=0.42 - 0.87 T`
(a) `phi = 1/2 a^(2)B=1/2 a^(2) (2+5t)`
`E=(d phi)/(dt) = 1/2 a^(2) xx 5`
`E=-(d phi)/(dt) = 1/2 a^(2) xx 5`
`E=10V`

Direction of induced emf will be in the same direction as the emf of battery.
`implies E_(n et)=E+20`
`:. E=30 V`
(b) Direction of current through the loop is anticlockwise.
`R_(n et)=(10+20) Omega = 30 Omega`
`:. I=(E_(n et))/(R_(n et)) = 1A`.