Question

The coil of an a.c generator rotates at a frequency of $60\text{Hz}$ and develops an induced emf of $120\text{V}$(rms). Area of the coil is $3\times {10}^{-3}{\text{m}}^2$ and number of turns is $N=500.$ The magnitude of magnetic field in which the coil rotates is

• A. $0.3\text{T}$
• B. $3\text{T}$
• C. $0.03\text{T}$
• D. $30\text{T}$

Answer 1

The correct option is A $0.3\text{T}$

Given,

$E_{rms}=120\text{V};\nu =60\text{Hz};A=3\times {10}^{-3}{\text{m}}^2;N=500$

We know that,

$E_{rms}=\frac{E_0}{\sqrt{2}}\Rightarrow E_0=\sqrt{2}{E}_{rms}$

$E_0=1.414\times 120=169.7\text{V}......\left(1\right)$

As, induced emf by the AC generator is,

$E=-\frac{Nd\varphi }{dt}=-N\frac{d}{dt}\left(BA\cos \theta \right)$

$=-N\frac{d}{dt}\left(BA\cos \omega t\right)[\because \theta =\omega t]$

$=NBA\omega \sin \omega t=E_0\sin \omega t$

$\Rightarrow E_0=NBA\omega ......\left(2\right)$

From (1) and (2),

$NBA\omega =169.7\text{V}$

$\Rightarrow B=\frac{169.7}{NA\omega }$

$=\frac{169.7}{500\times (3\times {10}^{-3})\times (2\pi \times 60)}\approx 0.3\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.