NEET UG 2017 — Physics Electromagnetism

2017mcqhard

A long solenoid of diameter $0.1m$ has $2\times {10}^{4}$ turns per meter. At the centre of the solenoid, a coil of $100$ turns and radius $0.01m$ is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to $0A$ from $4A$ in $0.05s$ . If the resistance of the coil is $10{\pi }^{2} \Omega ,$ the total charge flowing through the coil during this time is

Official previous-year question

Held on 30 Apr 2017 · Verified 9 Jul 2026.

Options

A
$32\pi \mu C$
B
$16 \mu C$
C
$32 \mu C$
D
$16\pi \mu C$

Did you get this right?

Solution

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:

$\epsilon = -N\frac{d\phi }{dt}$

$|\frac{\epsilon }{R}|=\frac{N}{R}\frac{d\phi }{dt}$

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.

$dq=\frac{N}{R}d\phi$

$\Delta Q=\frac{N(\Delta \phi )}{R}$

$\Delta Q=\frac{\Delta {\phi }_{total}}{R}$

$=\frac{(NBA)}{R}$

$=\frac{ N{\text{μ}}_{0} ni\text{π}{r}^{2} }{R}$

Putting values

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

$\Delta Q=32 \mu C$

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Solution

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:

$\epsilon = -N\frac{d\phi }{dt}$

$|\frac{\epsilon }{R}|=\frac{N}{R}\frac{d\phi }{dt}$

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.

$dq=\frac{N}{R}d\phi$

$\Delta Q=\frac{N(\Delta \phi )}{R}$

$\Delta Q=\frac{\Delta {\phi }_{total}}{R}$

$=\frac{(NBA)}{R}$

$=\frac{ N{\text{μ}}_{0} ni\text{π}{r}^{2} }{R}$

Putting values

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

$\Delta Q=32 \mu C$

NEET UG 2017 — Physics Electromagnetism

Options

Did you get this right?

Solution

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NEET UG 2017 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism