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A 800 turn coil of effective area 0.05 0.05 0.05 m 2 ^2 2 is kept perpendicular to a magnetic field 5 Ć 10 ā 5 5 \times 10^{-5} 5 Ć 1 0 ā 5 T . . . When the plane of the coil is rotated by 90 ā 90^\circ 9 0 ā around any of its coplanar axis in 0.1 0.1 0.1 s, the emf induced in the coil will be:
medium
Electromagnetic Induction
2019
physics
2 Ć 10 ā 3 2 \times 10^{-3} 2 Ć 1 0 ā 3 V
Explanation To solve this problem, we will use Faraday's law of electromagnetic induction. \\
The induced emf ( Īµ ) (\varepsilon) ( Īµ ) in a coil is given by: Īµ = ā N d Ī¦ d t \\
\varepsilon = -N \frac{d\Phi}{dt} \\
Īµ = ā N d t d Ī¦ ā
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V
where:
ā¢
is the number of turns in the coil.
ā¢
is the magnetic flux through one turn of the coil.
ā¢
d Ī¦ d t \frac{d\Phi}{dt} d t d Ī¦ ā is the rate of change of magnetic flux.
The magnetic flux
through the coil is given by:
Ī¦ = B ā
A ā
cos ā” Īø \\
\Phi = B \cdot A \cdot \cos \theta \\
Ī¦ = B ā
A ā
cos Īø where:
ā¢
B = 5 Ć 10 ā 5 B = 5 \times 10^{-5} B = 5 Ć 1 0 ā 5 T is the magnetic field strength.
ā¢
m
is the effective area of the coil.
ā¢
is the angle between the magnetic field and the normal to the coil.
Initially, the coil is perpendicular to the magnetic field, so
Īø = 0 ā . \theta = 0^\circ. \\
Īø = 0 ā . Thus, the initial magnetic flux
is:
Ī¦ 1 = B ā
A ā
cos ā” 0 ā = B ā
A \\
\Phi_1 = B \cdot A \cdot \cos 0^\circ = B \cdot A \\
Ī¦ 1 ā = B ā
A ā
cos 0 ā = B ā
A When the coil is rotated by
the angle
becomes
90 ā . 90^\circ. \\
9 0 ā . Thus, the final magnetic flux
is:
Ī¦ 2 = B ā
A ā
cos ā” 90 ā = 0 \\
\Phi_2 = B \cdot A \cdot \cos 90^\circ = 0 \\
Ī¦ 2 ā = B ā
A ā
cos 9 0 ā = 0 The change in magnetic flux
( Ī Ī¦ ) (\Delta \Phi) ( ĪĪ¦ ) is:
Ī Ī¦ = Ī¦ 2 ā Ī¦ 1 = 0 ā B ā
A = ā B ā
A \\
\Delta \Phi = \Phi_2 - \Phi_1 = 0 - B \cdot A = -B \cdot A \\
ĪĪ¦ = Ī¦ 2 ā ā Ī¦ 1 ā = 0 ā B ā
A = ā B ā
A The rate of change of magnetic flux
( d Ī¦ d t ) \left(\frac{d\Phi}{dt}\right) ( d t d Ī¦ ā ) is:
d Ī¦ d t = Ī Ī¦ Ī t = ā B ā
A 0.1 \\
\frac{d\Phi}{dt} = \frac{\Delta \Phi}{\Delta t} = \frac{-B \cdot A}{0.1} \\
d t d Ī¦ ā = Ī t ĪĪ¦ ā = 0.1 ā B ā
A ā Substitute the values:
d Ī¦ d t = ā ( 5 Ć 10 ā 5 ) ā
0.05 0.1 d Ī¦ d t = ā 2.5 Ć 10 ā 5 \\
\frac{d\Phi}{dt} = \frac{-(5 \times 10^{-5}) \cdot 0.05}{0.1} \\
\frac{d\Phi}{dt} = -2.5 \times 10^{-5} d t d Ī¦ ā = 0.1 ā ( 5 Ć 1 0 ā 5 ) ā
0.05 ā d t d Ī¦ ā = ā 2.5 Ć 1 0 ā 5 Wb/s
The induced emf
is:
Īµ = ā N d Ī¦ d t = ā 800 Ć ( ā 2.5 Ć 10 ā 5 ) Īµ = 2 Ć 10 ā 2 \\
\varepsilon = -N \frac{d\Phi}{dt} = -800 \times (-2.5 \times 10^{-5}) \\
\varepsilon = 2 \times 10^{-2} Īµ = ā N d t d Ī¦ ā = ā 800 Ć ( ā 2.5 Ć 1 0 ā 5 ) Īµ = 2 Ć 1 0 ā 2 V
Therefore, the emf induced in the coil is
V.
This corresponds to Option 4.