- Exam
- JEE Advanced
- Year
- 2022
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Magnetic Effects of Current
JEE Advanced 2022 · Paper 1 · Physics · Magnetic Effects of Current
A small circular loop of area A and resistance R is fixed on a...
A small circular loop of area and resistance is fixed on a horizontal -plane with the center of the loop always on the axis \overset{^}{n} of a long solenoid. The solenoid has turns per unit length and carries current counter clockwise as shown in the figure. The magnetic field due to the solenoid is in \overset{^}{n} direction. List-I gives time dependences of \overset{^}{n} in terms of a constant angular frequency . List-II gives the torques experienced by the circular loop at time , Let .
| List-I | List-II | ||
| (i) | \dfrac{1}{\sqrt{2}}\left(sin\omegat\overset{^}{j} + cos\omegat\overset{^}{k}\right) | (p) | |
| (ii) | \dfrac{1}{\sqrt{2}}\left(sin\omegat\overset{^}{i} + cos\omegat\overset{^}{j}\right) | (q) | - \dfrac{\alpha}{4}\overset{^}{i} |
| (iii) | \dfrac{1}{\sqrt{2}}\left(sin\omegat\overset{^}{i} + cos\omegat\overset{^}{k}\right) | (r) | \dfrac{3\alpha}{4}\overset{^}{i} |
| (iv) | \dfrac{1}{\sqrt{2}}\left(cos\omegat\overset{^}{j} + sin\omegat\overset{^}{k}\right) | (s) | \dfrac{\alpha}{4}\overset{^}{j} |
| (t) | - \dfrac{3\alpha}{4}\overset{^}{i} |
Which one of the following options is correct?
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
Note: This question was given bonus by JEE council.
The magnetic field in a solenoid is given by, \overset{\to}{B} = {\mu}_{0}mI\overset{^}{n}. Therefore, the flux through the loop will be,
ϕ = \left(A\overset{^}{k}\right) \cdot \left({\mu}_{0}mI\overset{^}{n}\right)
For case I:
The emf will be,
Therefore, the current in the loop will be,
The magnetic moment of the loop will be,
\overset{\to}{m} = iA\overset{^}{k} = \dfrac{B{A}^{2}\omega}{\sqrt{2}R}\sin \left(\omegat\right)\overset{^}{k}
Therefore, the torque
\overset{\to}{τ} = \overset{\to}{m} \times \overset{\to}{B} = \dfrac{{B}^{2}{A}^{2}\omega}{\sqrt{2}R}\sin \left(\omegat\right)\left(\overset{^}{k} \times \overset{^}{n}\right)
⇒ \overset{\to}{τ} = - \dfrac{{B}^{2}{A}^{2}\omega}{2R}\left[\overset{^}{i}\right]\sin^{2}\left(\omegat\right)
⇒ \overset{\to}{τ} = - \dfrac{{B}^{2}{A}^{2}\omega}{2R}\left[\sin^{2}\left(\dfrac{\pi}{6}\right)\right] = \dfrac{- \alpha}{4}\overset{^}{i}
Hence, \left(I\right) \to qϕ = 0τ = 0
Following the same process as in case I for case III
\overset{\to}{m} = \dfrac{B{A}^{2}\omega}{\sqrt{2}R}\sin \left(\omegat\right)\overset{^}{k}
\overset{\to}{τ} = \overset{\to}{m} \times \overset{\to}{B} = \dfrac{{B}^{2}{A}^{2}\omega}{\sqrt{2} \times \sqrt{2}R}sin\omegat\left(\overset{^}{k} \times \left(sin\omegat\overset{^}{i} + cos\omegat\overset{^}{k}\right)\right)
\overset{\to}{τ} = \dfrac{{B}^{2}{A}^{2}\omega sin\left(\omegat\right)}{2R}\sin \left(\omegat\right)\overset{^}{j}
\overset{\to}{τ} = \dfrac{{B}^{2}{A}^{2}\omega}{2R}\sin^{2}\left(\omegat\right)\overset{^}{j}
\overset{\to}{τ} = \dfrac{\alpha}{4}\overset{^}{j}
Hence,
Following the same process as in case I for case IV
\overset{\to}{m} = - \dfrac{B{A}^{2}\omega}{\sqrt{2}R}cos\omegat\left(\overset{^}{k}\right)
\overset{\to}{τ} = \overset{\to}{m} \times \overset{\to}{B} = - \dfrac{{B}^{2}{A}^{2}\omega}{2R}\left(\overset{^}{k} \times \overset{^}{j}\right)\cos^{2}\left(\omegat\right)
\overset{\to}{τ} = + \dfrac{{B}^{2}{A}^{2}\omega}{2R}\left(\overset{^}{i}\right) \cdot \cos^{2}\left(\dfrac{\pi}{6}\right)
\overset{\to}{τ} = + \dfrac{3}{4}\alpha\overset{^}{i}
Hence, \left(IV\right) \to r