Question

An ideal a solenoid of cross-sectional area
10-4 m² has 500 turns per meter. At the centre of
this solenoid, another coil of 100 turns is
wrapped closely around it. If the current in the
coil changes from 0 to 2 A in 3.14 millisecond,
the emf developed in the second coil is
(1) 1 mV (2) 2 mV (3) 3 mV (4) 4 mV​

Answer 1

Answer:

Option (4) 4 mV is correct.

Explanation:

Assumptions and Given values:

• $A$ = cross sectional area of the solenoid = $10^{-4}\ m^2$.

• $n$ = number of turns of solenoid = $500$ turns/m.

• $N$ = number of turns of the coil = $100$.

• $I_i$ = initial current = $0\ A$.

• $I_f$ = final current = $2\ A$.

• $dt$ = time interval in which the current is changed = $3.14\ ms = 3.14\times 10^{-3}\ s$.

The magnetic field produced by the solenoid is given by

$B=\mu_onI$

where, $\mu_o$ is the magnetic permeability of the free space, has value $\mu_o = 4\pi \times 10^{-7}\ N/A^2$.

The magnetic flux linked with the coil is given by

$\phi = NBA = N\mu_o nI\ A$

According to Faraday's law of electromagnetic induction, the emf induced in the second coil is given by

$e=-\dfrac{\mathrm{d} \phi}{\mathrm{d} t}=-\dfrac{\mathrm{d} (\mu_onI\ NA)}{\mathrm{d} t}  = -\mu_o nNA\dfrac{\mathrm{d} I}{\mathrm{d} t}$

$dI$ is the change in the current in time $dt$,

$dI = I_f-I_i = 2-0=2\ A.$

Putting all the values in the expression of the induced emf,

$e=-(4\pi \times 10^{-7}) \times (100)\times (500)\times (10^{-4})\times \dfrac{2}{3.14\times 10^{-3}} = - 4.002\times 10^{-3}\ V =- 4\ mV.$

The negative sign indicates that the emf is induced such that it opposes the change in flux through the coil.

Thus, option (4) is correct.