A coil of area A and N turns is rotating with angular velocity \( \omega\) in a uniform magnetic field \(\vec{B}\) about an axis perpendicular to \( \vec{B}\) Magnetic flux \(\varphi \text{ and induced emf } \varepsilon \text{ across it, at an instant when } \vec{B} \text{ is parallel to the plane of the coil, are:}\)
A coil of area A and N turns is rotating with angular velocity \( \omega\) in a uniform magnetic field \(\vec{B}\) about an axis perpendicular to \( \vec{B}\) Magnetic flux \(\varphi \text{ and induced emf } \varepsilon \text{ across it, at an instant when } \vec{B} \text{ is parallel to the plane of the coil, are:}\)
Show Hint
- \( \phi = AB, \varepsilon = 0 \)
- \( \phi = 0, \varepsilon = 0 \)
- \( \phi = 0, \varepsilon = NAB\omega \)
$\varphi = 0, \quad \varepsilon = N A B \omega$
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to understand the concept of electromagnetic induction in a rotating coil in a magnetic field.
Given:
- A coil with area \( A \) and \( N \) turns, rotating with angular velocity \( \omega \) in a uniform magnetic field \( \vec{B} \).
- The axis of rotation is perpendicular to the magnetic field \( \vec{B} \).
- We are asked to find the magnetic flux \( \varphi \) and the induced electromotive force (emf) \( \varepsilon \) when \( \vec{B} \) is parallel to the plane of the coil.
Key Concepts:
- Magnetic flux (\( \varphi \)) through a coil is given by the formula: \(\varphi = B \cdot A \cdot \cos \theta\), where \( \theta \) is the angle between the magnetic field \( \vec{B} \) and the normal to the plane of the coil.
- Induced emf (\( \varepsilon \)) is given by Faraday's law of electromagnetic induction: \(\varepsilon = -N \frac{d\varphi}{dt}\).
Analysis:
- When the magnetic field \( \vec{B} \) is parallel to the plane of the coil, the angle \( \theta = 90^\circ \), thus: \(\cos \theta = 0\).
- Consequently, the magnetic flux is: \(\varphi = B \cdot A \cdot \cos 90^\circ = 0\).
Calculating the induced emf (\( \varepsilon \)):
- Since the coil is rotating, the angle \( \theta \) will change with time, affecting the rate of change of flux.
- The expression for the emf, using that the coil rotates with angular velocity \( \omega \), becomes: \(\varepsilon = NAB\omega \cdot \sin(\omega t)\).
- At the specified instant, \(\theta = 90^\circ\), which leads to \(\cos \theta = 0\), so the rate of change of flux is maximal, thereby inducing the maximum emf: \(\varepsilon = NAB\omega\).
Conclusion:
- At the instant when \( \vec{B} \) is parallel to the plane of the coil, the magnetic flux is zero (since the magnetic field lines are parallel to the coil and not passing through it).
- The maximum emf is induced due to the change in flux related to the rotation of the coil, resulting in \( \varepsilon = NAB\omega \).
Thus, the correct answer is: \(\varphi = 0, \quad \varepsilon = NAB\omega\).
Approach Solution -2
The problem involves finding the magnetic flux \(\varphi\) and the induced electromotive force (emf) \(\varepsilon\) in a rotating coil within a magnetic field.
Here's a step-by-step analysis:
Magnetic Flux \(\varphi\):
The magnetic flux through a coil is given by \(\varphi = B \cdot A \cdot \cos(\theta)\), where \(B\) is the magnetic field, \(A\) is the area of the coil, and \(\theta\) is the angle between the magnetic field and the normal to the plane of the coil.
When the magnetic field \(\vec{B}\) is parallel to the plane of the coil, \(\theta = 90^\circ\), hence, the flux \(\varphi = B \cdot A \cdot \cos(90^\circ) = 0\).
Induced emf \(\varepsilon\):
The induced emf in a rotating coil is given by Faraday’s law of electromagnetic induction, \(\varepsilon = -\frac{d\varphi}{dt}\).
For a coil with area \(A\), \(N\) turns, and rotating with angular velocity \(\omega\) in a magnetic field \(B\), when the magnetic field is parallel to the plane of the coil, the rate of change of flux is maximal.
The flux \(\varphi = B \cdot A \cdot \cos(\theta) = B \cdot A \cdot \cos(\omega t)\).
Differentiating with respect to time \(t\), we get \(\varepsilon = -N \frac{d}{dt}(BA\cos(\omega t)) = NAB\omega\sin(\omega t)\).
At the instant when \(\theta = 90^\circ\), \(\sin(\omega t) = 1\). Thus, \(\varepsilon = NAB\omega\).
Thus, at the moment \(\vec{B}\) is parallel to the coil plane, we have:
Result:
\(\varphi = 0, \quad \varepsilon = NAB\omega\)
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