A long solenoid of diameter 0 . 1 m has 2 × 10 4 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0 . 01 m is placed with its axis coinciding with the solenoid axis. The

Question

A long solenoid of diameter 0 . 1 m has 2 × 10 4 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0 . 01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0 . 05 s . If the resistance of the coil is 10 π 2 Ω , the total charge flowing through the coil during this time isNEET 2017

TestHub · Accepted Answer

In the given case, the solenoid produces a changing magnetic field. This field is linked with the coil inside and hence, there is an induced emf in the coil.

We have, from the laws of EMI:

$ε = - N \frac{d ϕ}{d t}$

$|\frac{ε}{R}| = \frac{N}{R} \frac{d ϕ}{d t}$

The ratio given above is the induced current in the circuit which can further be written as the rate of flow of charge per unit time.

$d q = \frac{N}{R} d ϕ$

$Δ Q = \frac{N (Δ ϕ)}{R}$

$Δ Q = \frac{Δ ϕ_{t o t a l}}{R}$

$= \frac{(N B A)}{R}$

$= \frac{N μ_{0} n i π r^{2}}{R}$

Putting values

$= \frac{100 \times (4 π \times 10^{- 7}) \times (2 \times 10^{4}) \times 4 \times [{π \times (0.01)}^{2}]}{10 π^{2}}$

$Δ Q = 32 μC$

Physics - EMI/AC Question with Solution | TestHub

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