Physics Problem of the Day
Class XII — Electromagnetic Induction
A rectangular coil of 200 turns and area \(4 \times 10^{-2}\ m^2\) is rotated uniformly from a position where its plane is perpendicular to a magnetic field of strength \(0.5\ T\) to a position where its plane becomes parallel to the magnetic field in \(0.08\ s\). Calculate:
- The change in magnetic flux through the coil
- The average induced emf generated
Given:
- Number of turns, \(N = 200\)
- Area of coil, \(A = 4 \times 10^{-2}\ m^2\)
- Magnetic field, \(B = 0.5\ T\)
- Time taken, \(t = 0.08\ s\)
Detailed Solution
Step 1: Initial Magnetic Flux
Magnetic flux is given by:
\[ \Phi = BA\cos\theta \]
Initially, the plane of the coil is perpendicular to the magnetic field. Therefore, the normal to the coil is parallel to the field:
\[ \theta = 0^\circ \]
Hence:
\[ \Phi_i = BA\cos0^\circ \]
\[ \Phi_i = 0.5 \times 4\times10^{-2}\times1 \]
\[ \Phi_i = 2\times10^{-2}\ Wb \]
—Step 2: Final Magnetic Flux
Finally, the plane of the coil becomes parallel to the magnetic field. Therefore:
\[ \theta = 90^\circ \]
\[ \Phi_f = BA\cos90^\circ \]
\[ \Phi_f = 0 \]
—Step 3: Change in Flux
\[ \Delta\Phi = \Phi_f – \Phi_i \]
\[ \Delta\Phi = 0 – 2\times10^{-2} \]
\[ |\Delta\Phi| = 2\times10^{-2}\ Wb \]
—Step 4: Induced emf
According to Faraday’s Law:
\[ E = \frac{N|\Delta\Phi|}{t} \]
Substituting values:
\[ E = \frac{200 \times 2\times10^{-2}} {0.08} \]
\[ E = \frac{4}{0.08} \]
\[ E = 50\ V \]
—Final Answers
- Change in magnetic flux = \(2\times10^{-2}\ Wb\)
- Average induced emf = \(50\ V\)
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