Question

A small circular loop of area $A$ and resistance $R$ is fixed on a horizontal $x y$-plane with the center of the loop always on the axis $\hat{n}$ of a long solenoid The solenoid has $m$ turns per unit length and carries current $I$ counterclockwise as shown in the figure The magnetic field due to the solenoid is in $\hat{ n }$ direction List-I gives time dependences of $\hat{ n }$ in terms of a constant angular frequency $\omega$ List-II gives the torques experienced by the circular loop at time $t=\frac{\pi}{6 \omega}$, Let $\alpha=\frac{A^2 \mu_0^2 m^2 I^2 \omega}{2 R}$ 

Column I		Column II
I	$\frac{1}{\sqrt{2}}(\sin \omega t \hat{j}+\cos \omega t \hat{k})$	P	0
II	$\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{j})$	Q	$-\frac{\alpha}{4} \hat{i}$
III	$\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{k})$	R	$\frac{3 \alpha}{4} \hat{i}$
IV	$\frac{1}{\sqrt{2}}(\cos \omega t \hat{i}+\sin \omega t \hat{k})$	S	$\frac{\alpha}{4} \hat{j}$
		T	$-\frac{3 \alpha}{4} \hat{i}$

Which one of the following options is correct?

• A. $I \rightarrow Q , II \rightarrow P , III \rightarrow S , IV \rightarrow T$
• B. $I \rightarrow S , II \rightarrow T, III \rightarrow Q, IV \rightarrow P$
• C. $I \rightarrow Q , II \rightarrow P , III \rightarrow S , IV \rightarrow R$
• D. $I \rightarrow T , II \rightarrow Q , III \rightarrow P , IV \rightarrow R$

Answer 1

The Correct Option is C

Step 1: Understanding the Magnetic Field and Torque

We are dealing with a small circular loop of area \( A \) and resistance \( R \), placed in a solenoid carrying a time-varying current. The magnetic field inside the solenoid is given by \( B(t) = B_0 \cos(\omega t) \), where \( B_0 \) is the amplitude of the magnetic field, and \( \omega \) is the angular frequency.

The torque on the loop is given by: \[ \tau = \mu \times B \] where \( \mu = I A \hat{n} \) is the magnetic moment of the loop, and \( B \) is the magnetic field. The time-varying magnetic field induces a torque on the loop, which varies with time.

Step 2: Analyzing Column I Expressions

The expressions in Column I represent the time-varying magnetic field and the corresponding torques in Column II.

Expression I: \( \frac{1}{\sqrt{2}} (\sin(\omega t) + \cos(\omega t)) \)
This expression represents the time-varying magnetic field. Since both the sine and cosine terms oscillate with the same frequency, their combination results in a net zero average torque over one complete cycle. Therefore, the torque corresponding to this field is \({Q} \), which indicates zero torque.

Expression II: \( \frac{1}{\sqrt{2}} (\sin(\omega t) + \cos(\omega t)) \)
This expression is similar to Expression I. Given the same sinusoidal time-dependence, the net torque is again zero over one complete cycle, so the corresponding torque is \({P} \), which is also zero.

Expression III: \( \frac{1}{\sqrt{2}} (\sin(\omega t) + \cos(\omega t)) \)
This time-varying field produces a torque that is proportional to the angular frequency \( \omega \), which gives a non-zero torque that oscillates. The corresponding torque expression is \({S} \).

Expression IV: \( \frac{1}{\sqrt{2}} (\cos(\omega t) + \sin(\omega t)) \)
Like Expression III, this expression also results in a non-zero torque but with a different time-dependence. The corresponding torque expression for this is \({R} \).

Step 3: Final Answer

Based on the analysis, the correct matching is:

• Expression I corresponds to torque \( {Q} \) (zero torque),
• Expression II corresponds to torque \( {P} \) (zero torque),
• Expression III corresponds to torque \({S} \),
• Expression IV corresponds to torque \( {R} \).

Final Answer: The correct option is: C: I → Q, II → P, III → S, IV → R.