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 $\varphi \text{ and induced emf } \varepsilon \text{ across it, at an instant when } \vec{B} \text{ is parallel to the plane of the coil, are:}$

Question

A coil has $N$ turns and current passes through it is $I$ ampere then we obtain $L$ of self inductance. Now if current is made doubled then new self inductance be ______ H.

Answer 1

This problem requires understanding electromagnetic induction in a coil rotating within a magnetic field.

Parameters provided:

Coil characteristics: area $ A $, number of turns $ N $.
Rotation: angular velocity $ \omega $.
Magnetic field: uniform, denoted by $ \vec{B} $.
Rotation axis: perpendicular to $ \vec{B} $.
Objective: Determine magnetic flux $ \varphi $ and induced electromotive force (emf) $ \varepsilon $ when $ \vec{B} $ is parallel to the coil's plane.

Fundamental principles:

Magnetic flux ($ \varphi $) through a coil is calculated as: $\varphi = B \cdot A \cdot \cos \theta$. Here, $ \theta $ is the angle between the magnetic field $ \vec{B} $ and the normal to the coil's plane.
Induced emf ($ \varepsilon $) is governed by Faraday's law: $\varepsilon = -N \frac{d\varphi}{dt}$.

Analysis of conditions:

When $ \vec{B} $ is parallel to the coil's plane, the angle $ \theta = 90^\circ $, resulting in $\cos \theta = 0$.
Consequently, the magnetic flux is calculated as: $\varphi = B \cdot A \cdot \cos 90^\circ = 0$.

Calculation of induced emf ($ \varepsilon $):

The coil's rotation implies a time-varying angle $ \theta $, thus altering the rate of flux change.
With angular velocity $ \omega $, the emf expression is: $\varepsilon = NAB\omega \cdot \sin(\omega t)$.
At the specific instant where $ \theta = 90^\circ $ (and $ \cos \theta = 0 $), the rate of change of flux is at its maximum, inducing the maximum emf: $\varepsilon = NAB\omega$.

Summary of findings:

When $ \vec{B} $ is parallel to the coil's plane, the magnetic flux $ \varphi $ is zero because the magnetic field lines do not pass through the coil.
The rotation of the coil induces a maximum emf $ \varepsilon = NAB\omega $ due to the maximal rate of change of flux.

Therefore, the results are: $\varphi = 0, \quad \varepsilon = NAB\omega$.

Answer 2

This problem requires understanding electromagnetic induction in a coil rotating within a magnetic field.

Parameters provided:

Coil characteristics: area $ A $, number of turns $ N $.
Rotation: angular velocity $ \omega $.
Magnetic field: uniform, denoted by $ \vec{B} $.
Rotation axis: perpendicular to $ \vec{B} $.
Objective: Determine magnetic flux $ \varphi $ and induced electromotive force (emf) $ \varepsilon $ when $ \vec{B} $ is parallel to the coil's plane.

Fundamental principles:

Magnetic flux ($ \varphi $) through a coil is calculated as: $\varphi = B \cdot A \cdot \cos \theta$. Here, $ \theta $ is the angle between the magnetic field $ \vec{B} $ and the normal to the coil's plane.
Induced emf ($ \varepsilon $) is governed by Faraday's law: $\varepsilon = -N \frac{d\varphi}{dt}$.

Analysis of conditions:

When $ \vec{B} $ is parallel to the coil's plane, the angle $ \theta = 90^\circ $, resulting in $\cos \theta = 0$.
Consequently, the magnetic flux is calculated as: $\varphi = B \cdot A \cdot \cos 90^\circ = 0$.

Calculation of induced emf ($ \varepsilon $):

The coil's rotation implies a time-varying angle $ \theta $, thus altering the rate of flux change.
With angular velocity $ \omega $, the emf expression is: $\varepsilon = NAB\omega \cdot \sin(\omega t)$.
At the specific instant where $ \theta = 90^\circ $ (and $ \cos \theta = 0 $), the rate of change of flux is at its maximum, inducing the maximum emf: $\varepsilon = NAB\omega$.

Summary of findings:

When $ \vec{B} $ is parallel to the coil's plane, the magnetic flux $ \varphi $ is zero because the magnetic field lines do not pass through the coil.
The rotation of the coil induces a maximum emf $ \varepsilon = NAB\omega $ due to the maximal rate of change of flux.

Therefore, the results are: $\varphi = 0, \quad \varepsilon = NAB\omega$.

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The Correct Option is D

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