A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field B→ about an axis perpendicular to B→. Magnetic flux φ and induced emf ε across it, at an instant when B→ is parallel to the plane of coil, are :

Magnetic flux through the coil is given by:ϕ=NBAcos⁡θ\phi = N B A \cos\thetaSince the coil is rotating with angular velocity ω\omega, the angle at time tt is:θ=ωt\theta = \omega tThus, the flux varies as:ϕ=NBAcos⁡(ωt)\phi = N B A \cos(\omega t)At the instant when the magnetic field is parallel to the plane of the coil, the normal to the coil is perpendicular to the field. This means:θ=90∘ ⇒ cos⁡(90∘)=0\theta = 90^\circ \quad \Rightarrow \quad \cos(90^\circ) = 0 ϕ=NBA×0=0\phi = N B A \times 0 = 0According to Faraday's Law, the induced EMF is:ε=−dϕdt\varepsilon = - \frac{d\phi}{dt}Differentiating ϕ=NBAcos⁡(ωt)\phi = N B A \cos(\omega t):dϕdt=−NBAωsin⁡(ωt)\frac{d\phi}{dt} = - N B A \omega \sin(\omega t) ε=NBAωsin⁡(ωt)\varepsilon = N B A \omega \sin(\omega t)At the instant when BB is parallel to the plane of the coil, we already determined that θ=90∘\theta = 90^\circ (i.e., ωt=90∘\omega t = 90^\circ or π/2\pi/2).Since sin⁡90∘=1\sin 90^\circ = 1, we get:ε=NBAω\varepsilon = N B A \omega

A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field B→ about an axis perpendicular to B→. Magnetic flux φ and induced emf ε across it, at an instant when B→ is parallel to the plane of coil, are :

Magnetic flux through the coil is given by:ϕ=NBAcos⁡θ\phi = N B A \cos\thetaSince the coil is rotating with angular velocity ω\omega, the angle at time tt is:θ=ωt\theta = \omega tThus, the flux varies as:ϕ=NBAcos⁡(ωt)\phi = N B A \cos(\omega t)At the instant when the magnetic field is parallel to the plane of the coil, the normal to the coil is perpendicular to the field. This means:θ=90∘ ⇒ cos⁡(90∘)=0\theta = 90^\circ \quad \Rightarrow \quad \cos(90^\circ) = 0 ϕ=NBA×0=0\phi = N B A \times 0 = 0According to Faraday's Law, the induced EMF is:ε=−dϕdt\varepsilon = - \frac{d\phi}{dt}Differentiating ϕ=NBAcos⁡(ωt)\phi = N B A \cos(\omega t):dϕdt=−NBAωsin⁡(ωt)\frac{d\phi}{dt} = - N B A \omega \sin(\omega t) ε=NBAωsin⁡(ωt)\varepsilon = N B A \omega \sin(\omega t)At the instant when BB is parallel to the plane of the coil, we already determined that θ=90∘\theta = 90^\circ (i.e., ωt=90∘\omega t = 90^\circ or π/2\pi/2).Since sin⁡90∘=1\sin 90^\circ = 1, we get:ε=NBAω\varepsilon = N B A \omega

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A coil of area A and N turns is rotating with angular velocity $ω$ in a uniform magnetic field $\vec{B}$ about an axis perpendicular to $\vec{B}$ . Magnetic flux $φ$ and induced emf $ε$ across it, at an instant when $\vec{B}$ is parallel to the plane of coil, are :

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$φ$ = AB, $ε$ = NAB $ω$

$φ$ = 0, $ε$ = 0

$φ$ = 0, $ε$ = NAB $ω$

$φ$ = AB, $ε$ = 0

answer is B.

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Detailed Solution

Magnetic flux through the coil is given by:

$\phi = N B A \cos\theta$

Since the coil is rotating with angular velocity $\omega$ , the angle at time $t$ is:

$\theta = \omega t$

Thus, the flux varies as:

$\phi = N B A \cos(\omega t)$

At the instant when the magnetic field is parallel to the plane of the coil, the normal to the coil is perpendicular to the field. This means:

$\theta = 90^\circ \quad \Rightarrow \quad \cos(90^\circ) = 0$

$\phi = N B A \times 0 = 0$

According to Faraday's Law, the induced EMF is:

$\varepsilon = - \frac{d\phi}{dt}$

Differentiating $\phi = N B A \cos(\omega t)$ :

$\frac{d\phi}{dt} = - N B A \omega \sin(\omega t)$

$\varepsilon = N B A \omega \sin(\omega t)$

At the instant when $B$ is parallel to the plane of the coil, we already determined that $\theta = 90^\circ$ (i.e., $\omega t = 90^\circ$ or $\pi/2$ ).