Q169 / 180 · COMEDK 2024 (Morning Shift) · Physics
Around the central part of an air cored solenoid of length $20 \mathrm{~cm}$ and area of cross section $1.4 \times 10^{-3} \mathrm{~m}^2$ and 3000 turns, another coil of 250 turns is closely wound. A current $2 \mathrm{~A}$ in the solenoid is reversed in $0.2 \mathrm{~s}$, then the induced emf produced is
- Correct answerA$1.32 \times 10^{-1} \mathrm{~V}$
- B$1.16 \times 10^{-1} \mathrm{~V}$
- C$4 \times 10^{-1} \mathrm{~V}$
- D$8 \times 10^{-2} \mathrm{~V}$
Solution
The magnetic field $B$ inside a long solenoid is given by $B = \mu_{0} n I$, where $n = \dfrac{N}{l}$ is the number of turns per unit length.
Given: $N_{1} = 3000$, $l = 0.2 \text{ m}$, $I_{initial} = 2 \text{ A}$, $I_{final} = -2 \text{ A}$, $A = 1.4 \times 10^{-3} \text{ m}^2$, $N_{2} = 250$, $\Delta t = 0.2 \text{ s}$.
The magnetic flux $\phi$ through the secondary coil is $\phi = B A = \mu_{0} \left( \dfrac{N_{1}}{l} \right) I A$.
The change in magnetic flux $\Delta \phi$ is $\Delta \phi = \phi_{final} - \phi_{initial} = \mu_{0} \left( \dfrac{N_{1}}{l} \right) A (I_{final} - I_{initial})$.
Substituting the values:
$\Delta \phi = (4 \pi \times 10^{-7}) \times \left( \dfrac{3000}{0.2} \right) \times (1.4 \times 10^{-3}) \times (-2 - 2)$
$\Delta \phi = (4 \pi \times 10^{-7}) \times (15000) \times (1.4 \times 10^{-3}) \times (-4)$
$\Delta \phi = -1.055 \times 10^{-4} \text{ Wb}$ (approx).
The induced emf $\varepsilon$ in the secondary coil is given by Faraday's law: $\varepsilon = -N_{2} \dfrac{\Delta \phi}{\Delta t}$.
$\varepsilon = -250 \times \dfrac{-1.055 \times 10^{-4}}{0.2} = 1250 \times 1.055 \times 10^{-4} = 0.131875 \text{ V}$.
Rounding to the nearest value, $\varepsilon \approx 1.32 \times 10^{-1} \text{ V}$.
Answer: $1.32 \times 10^{-1} \mathrm{~V}$
Given: $N_{1} = 3000$, $l = 0.2 \text{ m}$, $I_{initial} = 2 \text{ A}$, $I_{final} = -2 \text{ A}$, $A = 1.4 \times 10^{-3} \text{ m}^2$, $N_{2} = 250$, $\Delta t = 0.2 \text{ s}$.
The magnetic flux $\phi$ through the secondary coil is $\phi = B A = \mu_{0} \left( \dfrac{N_{1}}{l} \right) I A$.
The change in magnetic flux $\Delta \phi$ is $\Delta \phi = \phi_{final} - \phi_{initial} = \mu_{0} \left( \dfrac{N_{1}}{l} \right) A (I_{final} - I_{initial})$.
Substituting the values:
$\Delta \phi = (4 \pi \times 10^{-7}) \times \left( \dfrac{3000}{0.2} \right) \times (1.4 \times 10^{-3}) \times (-2 - 2)$
$\Delta \phi = (4 \pi \times 10^{-7}) \times (15000) \times (1.4 \times 10^{-3}) \times (-4)$
$\Delta \phi = -1.055 \times 10^{-4} \text{ Wb}$ (approx).
The induced emf $\varepsilon$ in the secondary coil is given by Faraday's law: $\varepsilon = -N_{2} \dfrac{\Delta \phi}{\Delta t}$.
$\varepsilon = -250 \times \dfrac{-1.055 \times 10^{-4}}{0.2} = 1250 \times 1.055 \times 10^{-4} = 0.131875 \text{ V}$.
Rounding to the nearest value, $\varepsilon \approx 1.32 \times 10^{-1} \text{ V}$.
Answer: $1.32 \times 10^{-1} \mathrm{~V}$
Hence, the correct answer is option: 'A'.