A long solenoid having n = 200 turns per metre has a circular cross-section of radius `a_(1) = 1 cm`. A circular conducting loop of radius `a_(2) = 4 cm` and resistance ` R = 5 (Omega)` encircles the solenoid such that the centre of circular loop coincides with the midpoint of the axial line of the solenoid and they have the same axis as shown in Fig.

A current 't' in the solenoid results in magnetic field along its axis with magnitude `B = (mu)ni` at points well inside the solenoid on its axis. We can neglect the insignificant field outside the solenoid. This results in a magnetic flux `(phi)_(B)` through the circular loop. If the current in the winding of solenoid is changed, it will also change the magnetic field `B = (mu)_(0)ni` and hence also the magnetic flux through the circular loop. Obvisouly, it will result in an induced emf or induced electric field in the circular loop and an induced current will appear in the loop. Let current in the winding of solenoid be reduced at a rate of `75 A //sec`.
Magnitude of induced current that appears in the circular loop is

Options

A. `2.72 mu A`
B. `1.18 mu A`
C. `19.2 mu A`
D. None of these

Answer

B

Explanation

Here, `(dB)/(dt)= mu_(0)n(di)/(dt)`
Induced emf becomes `e=pi a_(1)^(2) (dB)/(dt)`
thus, induced current is
`i=e?R = (pi xx 10^(-4) xx 4 pi xx 10^(-7) xx 200 xx 75)/(5) implies 1.18 mu A`.