Question 52: Electromagnetic Induction

Question

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $I=10 \sin (\omega \mathrm{t}) \mathrm{A}$, where $\omega=1000 \mathrm{rad} . / \mathrm{s}$. A circular conducting loop $(B)$ of radius 1 cm coaxially slided through the solenoid at a speed $v=1 \mathrm{~cm} / \mathrm{s}$. The r.m.s. current through the loop when the coil $B$ is inserted 10 cm inside the solenoid is $${\alpha  \over {\sqrt 2 }}\mu A$$. The value of $\alpha$ is $\_\_\_\_$ .
[Resistance of the loop $=10 \Omega$ ]

Accepted Answer

For a long solenoid, the magnetic field B inside is uniform and given by :
$$ \mathrm{B}=\mu_0 \mathrm{nI} $$
Where :
$\mu_0=4 \pi 	imes 10^{-7} \mathrm{~T} \cdot \mathrm{~m} / \mathrm{A}$
$\mathrm{n}=500$ turns/unit length.
$\mathrm{I}=10 \sin (1000 \mathrm{t}) \mathrm{A}$
The loop is inside the solenoid, so it experiences this magnetic field. The flux through the circular loop of radius $\mathrm{r}=1 \mathrm{~cm}=0.01 \mathrm{~m}$ is :
$$ \Phi=\mathrm{B} \cdot \mathrm{~A}=\left(\mu_0 \mathrm{nI}ight) \cdot\left(\pi \mathrm{r}^2ight) $$
$$\Rightarrow $$ $\Phi=\mu_0 \mathrm{n}\left(\pi \mathrm{r}^2ight) \cdot 10 \sin (1000 \mathrm{t})$
According to Faraday's Law, the induced EMF is $\varepsilon=-\frac{\mathrm{d} \Phi}{\mathrm{dt}}$.
Since the loop is already 10 cm inside a 100 cm long solenoid, it is in the region where the magnetic field is uniform. When the loop is inside the magnetic field then its motion along axis of solenoid will not cause any change in effective magnetic flux through loop because loop is co-axial with the solenoid.
$$ \varepsilon=\frac{d}{d t}\left[\mu_0 n \pi r^2 \cdot 10 \sin (1000 t)ight] $$
$$\Rightarrow $$$$ \varepsilon=\mu_0 \mathrm{n} \pi \mathrm{r}^2 \cdot 10 \cdot 1000 \cos (1000 \mathrm{t}) $$
The induced current i is $\frac{\varepsilon}{\mathrm{R}}$ :
$$ \mathrm{i}=\frac{10000 \mu_0 \mathrm{n} \pi \mathrm{r}^2}{\mathrm{R}} \cos (1000 \mathrm{t}) $$
Peak induced current $\mathrm{i}_0=\frac{10^4 	imes\left(4 \pi 	imes 10^{-7}ight) 	imes 500 	imes \pi 	imes(0.01)^2}{10} \mathrm{~A}$
$$\Rightarrow $$ $$ \mathrm{i}_0 \approx 1.97 	imes^{-4} \mathrm{~A}=197 	imes 10^{-6} \mathrm{~A}=197 \mu \mathrm{~A} $$
RMS current of a sinusoidal current is given as,
$$ \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_0}{\sqrt{2}}=\frac{197}{\sqrt{2}} \mu \mathrm{~A}=\frac{\alpha}{\sqrt{2}} \mu \mathrm{~A} $$
$\Rightarrow \alpha=197$

Answer

80

Answer

280

Answer

100

Answer

197

Electromagnetic Induction

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