Question

A coil having 200 turns has a surface area 0.12 m². A magnetic field of strength 0.10 Wbm⁻² linked perpendicular to this area changes to 0.5 Wbrn⁻² in 0.2 s. Find the average emf induced in the coil. [Ans:48 V]

Answer 1

Answer:

Average value of induced emf = 48 V.

Explanation:

Given:

Number of turns of the coil, $N = 200$.

Surface area of the coil, $A = 0.12\ m^2$.

Strength of the magnetic field initially, $B_i = 0.10\ Wbm^{-2}$.

Strength of the magnetic field after 0.2 s, $B_f=0.5\ Wbm^{-2}$.

The magnetic flux through a surface is defined as

$\Phi = N\vec B \cdot \vec A = NBA\cos \theta$

where,

$\vec A$ = area vector of the coil which is directed along the normal to the plane of the coil and has magnitude equal to the surface area of the coil.

$\theta$ = angle between $\vec B$ and $\vec A$.

In the given question, the magnetic field is perpendicular to the area of the coil which means it is along the same direction as that of $\vec A$.

Therefore, $\theta = 0^\circ$.

The magnetic flux linked with the coil is given as

$\Phi = NBA\cos 0^\circ = NBA.$

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

$e=-\dfrac{\mathrm{d} \Phi}{\mathrm{d} t}\\=-\dfrac{\mathrm{d} (NBA)}{\mathrm{d} t}\\=-(NA)\dfrac{\mathrm{d} B}{\mathrm{d} t}$

$\mathrm dB$ is the change in the magnetic field = $B_f-B_i = 0.5-0.10= 0.40\ Wbm^{-2}.\\\mathrm dt = 0.2\ s.$

Therefore,

$e= -(200)\times (0.12)\times \dfrac{0.40}{0.2}\\=-48\ V.$

The average induced emf is given by

$e_{avg} = \dfrac{\int e\ dt}{\int dt}$

Since, there is no time dependent term in the value of the induced emf, therefore,

$e_{avg} = \dfrac{e\int \ dt}{\int dt}= e$

Thus, the magnitude of average emf induced in the coil is same as the magnitude of induced emf i.e., 48 V.

Answer 2

Dear Student,

◆ Answer -

EMF = -48 V

◆ Explanation -

# Given -

N = 200 turns

B1 = 0.1 Wb/m^2

B2 = 0.5 Wb/m^2

A = 0.12 m^2

t = 0.2 s

# Solution -

EMF induced in the coil is given by -

EMF = -∆Φ/t

EMF = -N.(B₂-B₁).A/t

EMF = -200 × (0.5 - 0.1) × 0.12 / 0.2

EMF = -48 V

Therefore, magnitude of average emf induced in the coil is 48 V.

Hope this helps you. Thanks dear...