Physics 3 Marks Question

$\Rightarrow\mu_0\text{ni}=\mu\text{n }\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\text{t}}\Big)$

$\Rightarrow\text{B}=\text{B}_0\Big(1-\text{e}^\frac{-\text{lR}}{\text{L}}\Big)$

$\Rightarrow0.8\text{ B}_0=\text{B}_0\Big(1-\text{e}^\frac{-20\times10^{-5}\times\text{R}}{2\times10^{-3}}\Big)$

$\Rightarrow0.8=\Big(1-\text{e}^\frac{-\text{R}}{100}\Big)$

$\Rightarrow\text{e}^\frac{-\text{R}}{100}=0.2$

$\Rightarrow\text{ln}\Big(\text{}e^\frac{-\text{R}}{100}\Big)=\text{ln}\big(0. 2\big)$

$\Rightarrow\frac{-\text{R}}{100}=-1.609$

$\Rightarrow\text{R}=16.9=160\Omega.$

$\text{dl}=\text{l}_2-\text{l}_1=2.5-\big(-2.5\big)=5\text{A} $

$\text{dt}=0.1\text{s}$

$\text{V}=\text{L}\frac{\text{dl}}{\text{dt}}$

$\Rightarrow20=\text{L}\Big(\frac{5}{0.1}\Big)\Rightarrow20=\text{L}\times50$

$\Rightarrow\text{L}=\frac{20}{50}=\frac{4}{10}=0.4\text{ Henry}$

$\phi$ througth the coil

$\phi=\text{N.B.A}$

$\phi=\text{N.}\mu_0\text{ni}\pi\text{R}^2$

1. $\text{e}=\frac{\text{d}\phi}{\text{dt}}$

$=\frac{\text{d}}{\text{dt}}\big(\mu_0\text{n}\text{N}\pi\text{R}^2\text{i}\big)$

$=\mu_0\text{nN}\pi\text{R}^2\frac{\text{d}}{\text{dt}}(\text{i}_0\sin\omega\text{t})$

$=\mu_0\text{nN}\pi\text{R}^2\text{i}_0\cos\omega\text{t}\times\omega$

$\text{e}=\mu_0\text{nN}\pi\text{R}^2\text{i}_0\omega\cos\omega\text{t}$

2. $\text{E}=\text{M}\frac{\text{d}\text{I}}{\text{dt}}$

$\text{E}=\mu_0\text{nN}\pi\text{R}^2\frac{\text{d}\text{i}}{\text{dt}}$

$\text{M}=\mu_0\text{nN}\pi\text{R}^2$ 

1. $\text{i}_{0} = \frac{\text{e}}{\text{R}}=\frac{4}{20},\tau=\frac{\text{L}}{\text{R}}=\frac{2}{20}=0.1 $

$ \text{i}=\text{i}_{0}(1-\text{e}^{\frac{-t}{\tau}})=\frac{4}{20}\Big(1-e^\frac{-0.2}{0.1}\Big) $

$= 0 .17\text{A}$

2. $ \frac{1}{2}\text{Li}^2 =\frac{1}{2} \times2\times(0.17)^2=0.0289=0.03\text{J}$

$\frac {\text{dB}}{\text{dt}}=0.010\text{T}/',\frac{\text{d}\phi}{\text{dt}}=\frac{\text{dB}}{\text{dt}}\text{A}$

$\text{E}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{dB}}{\text{dt}}\times\text{A}=0.01\Big(\frac{\pi\times\text{r}^2}{2}\Big)$

$=\frac{0.01\times3.14\times0.01}{2}=\frac {3.14}{2}\times10^{-4}=1.57\times10^{-4}$

$\text{i}=\frac{\text{E}}{\text{R}}=\frac{1.57\times10^{-4}}{4}=0.39\times10^{-4}=0.39\times10^{-5}\text{A}$

$\text{n}=200,\text{l}=4\text{A},\text{E}=-\text{nl}\frac{\text{dl}}{\text{dt}}$

$\text{or},\frac{-\text{d}\phi}{\text{dt}}=\frac{-\text{Ldl}}{\text{dt}}$

$\text{or},\text{L}=\text{n}\frac{-\text{d}\phi}{\text{dt}}=200\times8\times10^{-4}=2\times2\times10^{-2}\text{H}.$

$=\frac{4\pi\times10^{-7}\times\big(240\big)^2\times\pi\big(2\times10^{-2}\big)^2}{12\times10^{-2}}\times0.8$

$\frac{4\pi\times\big(24\big)^2\times\pi\times4\times8}{12}\times10^{-8}$

$=60577.3824\times10^{-8}=6\times10^{-4}\text{V}$

$=\text{BA}\sin\theta=-\text{BA}\omega\sin\theta$

$\Big(\frac{\text{dq}}{\text{dt}}=$ the rate of change of angle between arc vector and $\text{B}=\omega\Big)$

1. For maximum emf, $\sin\theta=1$

$\therefore\text{e}=\text{BA}\omega$

$\Rightarrow\text{e}=0.010\times25\times10^{-4}\times80\times\frac{2\pi\times\pi}{60}$

$\Rightarrow\text{e}=0.66\times10^{-3}=6.66\times10^{-4}\text{V}$

2. The direction of the induced emf changes every instant. Thus, the average emf becomes zero.

3. The emf induced in the coil is $\text{e}=-\text{BA}\omega\sin\theta=-\text{BA}\omega\sin \ \omega\text{t}$

The average of the squares of emf induced is given by

$\text{e}_{\text{av}^2}=\frac{\int\limits_{0}^{\text{T}}\text{B}^2\text{A}^2\omega^2\sin^2\omega\text{t} \ \text{dt}}{\int\limits_{0}^{\text{T}}\text{dt}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2\int\limits_{0}^{\text{T}} \sin^2\omega\text{t} \ \text{dt}}{\int\limits_{0}^{\text{T}}\text{dt}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2\int\limits_{0}^{\text{T}} (1-\cos2\omega\text{t} )\ \text{dt}}{ \ 2\text{T}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2 }{ 2\text{T}}\Big[\frac{\text{t}-\sin2\omega\text{t}}{ \ 2\omega}\Big]_{0}^{\text{T}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2}{ \ 2\text{T}}\Big[\text{T}-\frac{\sin4\pi-\sin0}{2\omega}\Big]=\frac{\text{B}^2\text{A}^2\omega^2}{2}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{(6.66\times10^{-4})2}{2}$

$=22.1778\times10^{-8}\text{V}^2 \ \big[\because\text{BA}\omega=6.66\times10^{-4}\text{V}\big]$

$\Rightarrow\text{e}-{\text{av}^2}=2.2\times10^{-7}\text{V}^2$