Question 29: Electromagnetic Induction

Question

A 30 cm long solenoid has 10 turns per cm and area of $5 \mathrm{~cm}^2$. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s . The e.m.f. induced in the coil is $\alpha 	imes 10^{-5} \mathrm{~V}$. The value $\alpha$ is $\_\_\_\_$ .

Accepted Answer

According to Faraday's Law, a change in the magnetic flux $(\Phi)$ through a coil induces an electromotive force (e.m.f.) in that coil. The magnitude of the induced e.m.f. ( $\varepsilon$ ) is given by:
$$ \varepsilon=N \frac{d \Phi}{d t} $$
Where N is the total number of turns. For a solenoid, the magnetic flux $\Phi$ through each turn is $\Phi=\mathrm{B} \cdot \mathrm{A}$, where $B$ is the magnetic field inside the solenoid and $A$ is the cross-sectional area.
The magnetic field $B$ inside an ideal solenoid is:
$$ \mathrm{B}=\mu_0 \mathrm{nI} $$
Where $\mu_0$ is the permeability of free space $\left(4 \pi 	imes 10^{-7} \mathrm{~T} \cdot \frac{\mathrm{~m}}{\mathrm{~A}}ight)$, n is the number of turns per unit length and I is the current.
The total flux linkage is $\Psi=N \Phi=(n l)(B A)=(n l)\left(\mu_0 n I Aight)=\mu_0 n^2 l A I$.
Since $\varepsilon=\frac{\mathrm{d} \Psi}{\mathrm{dt}}$,
$$ \varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}} $$
Where the self-inductance L is $\mathrm{L}=\mu_0 \mathrm{n}^2$ lA
The length of the solenoid is $\mathrm{l}=30 \mathrm{~cm}=0.3 \mathrm{~m}$
Turns per $\mathrm{cm}=10$, so turns per meter will be $\mathrm{n}=10 	imes 100=1000$ turns $/ \mathrm{m}$.
The area of cross-section is $\mathrm{A}=5 \mathrm{~cm}^2=5 	imes 10^{-4} \mathrm{~m}^2$.
So,
$L=\left(4 	imes 3.14 	imes 10^{-7}ight) 	imes(1000)^2 	imes 0.3 	imes\left(5 	imes 10^{-4}ight)$
$$\Rightarrow $$ $\mathrm{L}=12.56 	imes 10^{-7} 	imes 10^6 	imes 0.3 	imes 5 	imes 10^{-4}$
$$\Rightarrow $$ $$ \mathrm{L}=12.56 	imes 1.5 	imes 10^{-5}=18.84 	imes 10^{-5} \mathrm{H} $$
The change in current is $\Delta \mathrm{I}=4 \mathrm{~A}-2 \mathrm{~A}=2 \mathrm{~A}$.
The time interval ( $\Delta \mathrm{t}$ ) is 3.14 s.
$$ \varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}} $$
$$\Rightarrow $$ $\varepsilon=\left(18.84 	imes 10^{-5}ight) 	imes \frac{2}{3.14}$
$$\Rightarrow $$ $\varepsilon=6 	imes 2 	imes 10^{-5}$
$$\Rightarrow $$ $$ \varepsilon=12 	imes 10^{-5} \mathrm{~V}=\alpha 	imes 10^{-5} \mathrm{~V} \Rightarrow \alpha=12 $$
Therefore, the value of $\alpha$ is 12 .
Hence, the correct option is (B).

Answer

60

Answer

12

Answer

120

Answer

34

Electromagnetic Induction

A 30 cm long solenoid has 10 turns per cm and area of 5 \mathrm{~cm}^2. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s . The e.m.f. induced in the coil is \alpha \times 10^{-5} \mathrm{~V}. The value \alpha is \_\_\_\_ .

Choose the correct answer: