28 days until NEET 2026 Every question counts. ✊
🧠
Did you know?
The human brain has ~86 billion neurons, each connected to up to 7,000 others.
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 × 10 − 5 5 \times 10^{-5} 5 × 1 0 − 5 T . . . When the plane of the coil is rotated by 90 ∘ 90^\circ 9 0 ∘ 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 × 10 − 3 2 \times 10^{-3} 2 × 1 0 − 3 V
Explanation To solve this problem, we will use Faraday's law of electromagnetic induction. \\
The induced emf ( ε ) (\varepsilon) ( ε ) in a coil is given by: ε = − N d Φ d t \\
\varepsilon = -N \frac{d\Phi}{dt} \\
ε = − N d t d Φ
Our AI powered practice platform can help you achieve your doctor dream. Practice 2000+ previous year NEET questions More practice, more score
AI generated hints Use hints to get start solving
AI companion chat to clear doubts 24*7 Ask any question, get instant answers
AI generated solutions Get detailed step by step solutions
Check related NCERT content Read while solving
Track your progress Improve every day
Sign up / Login
V
where:
•
is the number of turns in the coil.
•
is the magnetic flux through one turn of the coil.
•
d Φ d t \frac{d\Phi}{dt} d t d Φ is the rate of change of magnetic flux.
The magnetic flux
through the coil is given by:
Φ = B ⋅ A ⋅ cos θ \\
\Phi = B \cdot A \cdot \cos \theta \\
Φ = B ⋅ A ⋅ cos θ where:
•
B = 5 × 10 − 5 B = 5 \times 10^{-5} B = 5 × 1 0 − 5 T is the magnetic field strength.
•
m
is the effective area of the coil.
•
is the angle between the magnetic field and the normal to the coil.
Initially, the coil is perpendicular to the magnetic field, so
θ = 0 ∘ . \theta = 0^\circ. \\
θ = 0 ∘ . Thus, the initial magnetic flux
is:
Φ 1 = B ⋅ A ⋅ cos 0 ∘ = B ⋅ A \\
\Phi_1 = B \cdot A \cdot \cos 0^\circ = B \cdot A \\
Φ 1 = B ⋅ A ⋅ cos 0 ∘ = B ⋅ A When the coil is rotated by
the angle
becomes
Thus, the final magnetic flux
is:
Φ 2 = B ⋅ A ⋅ cos 90 ∘ = 0 \\
\Phi_2 = B \cdot A \cdot \cos 90^\circ = 0 \\
Φ 2 = B ⋅ A ⋅ cos 9 0 ∘ = 0 The change in magnetic flux
is:
Δ Φ = Φ 2 − Φ 1 = 0 − B ⋅ A = − B ⋅ A \\
\Delta \Phi = \Phi_2 - \Phi_1 = 0 - B \cdot A = -B \cdot A \\
ΔΦ = Φ 2 − Φ 1 = 0 − B ⋅ A = − B ⋅ A The rate of change of magnetic flux
( d Φ d t ) \left(\frac{d\Phi}{dt}\right) ( d t d Φ ) is:
d Φ d t = Δ Φ Δ t = − B ⋅ A 0.1 \\
\frac{d\Phi}{dt} = \frac{\Delta \Phi}{\Delta t} = \frac{-B \cdot A}{0.1} \\
d t d Φ = Δ t ΔΦ = 0.1 − B ⋅ A Substitute the values:
d Φ d t = − ( 5 × 10 − 5 ) ⋅ 0.05 0.1 d Φ d t = − 2.5 × 10 − 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 Φ = 0.1 − ( 5 × 1 0 − 5 ) ⋅ 0.05 d t d Φ = − 2.5 × 1 0 − 5 Wb/s
The induced emf
is:
ε = − N d Φ d t = − 800 × ( − 2.5 × 10 − 5 ) ε = 2 × 10 − 2 \\
\varepsilon = -N \frac{d\Phi}{dt} = -800 \times (-2.5 \times 10^{-5}) \\
\varepsilon = 2 \times 10^{-2} ε = − N d t d Φ = − 800 × ( − 2.5 × 1 0 − 5 ) ε = 2 × 1 0 − 2 V
Therefore, the emf induced in the coil is
V.
This corresponds to Option 4.