A long solenoid with 2000 turns per meter has a small loop of radius 3cm placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from 1.5 A to 5.5 A in `(pi^(2))/(100)` s, the induced emf in the loop is

Options

A. 0.144 mV
B. 0.288 mV
C. 0.072 mV
D. 0.316 mV

Answer

B

Explanation

Magnetic field inside a solenoid of infinite length is given by expression
`B = mu_(0) n i`
where, n = number of turns per unit length.
Given. Number of turns in solenoid, n = 2000, current through solenoid, `i_(i) = 1.5 A` and `i_(f) = 5.5 A`
So, `Delta B = mu_(0) n(i_(f) - i_(i))`
Putting the given values,
`= 4pi xx 10^(-7) xx 2000 (5.5 - 1.5)`
`rArr Delta B = 4pi xx 10^(-7) xx 2000 xx 4`
Now, emf induced in a loop of radius 3 cm,
`e = -(d phi)/(d t) = (Delta BA)/(Delta t) cos theta`
`rArr e = (4pi xx 10^(-7) xx 2000 xx 4 xx pi xx (3 xx 10^(-2))^(2)(cos 0^(@)))/((pi^(2))/(100))`
`rArr E = 0.288 mV " " (because theta = 0^(@))`
Hence, the correct option is (b).`
`rArr E = 0.288 mV " " (because theta = 0^(@))`
Hence, the correct option is (b).