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 n ^ of a long solenoid. The solenoid has m turns per unit length and carries current I counter clockwise as shown in the figure. The magnetic field due to the…

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 n ^ of a long solenoid. The solenoid has m turns per unit length and carries current I counter clockwise as shown in the figure. The magnetic field due to the solenoid is in n ^ direction. List-I gives time dependences of n ^ in terms of a constant angular frequency &#969; . List-II gives the torques experienced by the circular loop at time t = &#960; 6 &#969; , Let &#945; = A 2 &#956; 0 2 m 2 I 2 &#969; 2 R . List-I List-II (i) 1 2 sin &#969; t j ^ + cos &#969; t k ^ (p) 0 (ii) 1 2 sin &#969; t i ^ + cos &#969; t j ^ (q) - &#945; 4 i ^ (iii) 1 2 sin &#969; t i ^ + cos &#969; t k ^ (r) 3 &#945; 4 i ^ (iv) 1 2 cos &#969; t j ^ + sin &#969; t k ^ (s) &#945; 4 j ^ (t) - 3 &#945; 4 i ^ Which one of the following options is correct?

Aicharya · Accepted Answer

Note: This question was given bonus by JEE council.
The magnetic field in a solenoid is given by, $\vec{B} = μ_{0} m I \hat{n}$ . Therefore, the flux through the loop will be,
$ϕ = (A \hat{k}) \cdot (μ_{0} m I \hat{n})$
For case I:
$ϕ = \frac{B A}{\sqrt{2}} \cos (ω t)$
The emf will be,
$ε = \frac{B A ω}{\sqrt{2}} \sin (ω t)$
Therefore, the current in the loop will be,
$i = \frac{B A ω}{\sqrt{2} R} \sin (ω t)$
The magnetic moment of the loop will be,
$\vec{m} = i A \hat{k} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k}$
Therefore, the torque
$\vec{τ} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{\sqrt{2} R} \sin (ω t) (\hat{k} \times \hat{n})$
$\Rightarrow \vec{τ} = - \frac{B^{2} A^{2} ω}{2 R} [\hat{i}] \sin^{2} (ω t)$
$\Rightarrow \vec{τ} = - \frac{B^{2} A^{2} ω}{2 R} [\sin^{2} (\frac{π}{6})] = \frac{- α}{4} \hat{i}$
Hence, $(I) \to q$
For case II
$ϕ = 0$
Therefore, we can directly write $τ = 0$
Hence $(II) \to p$
Following the same process as in case I for case III
$ϕ = \frac{B A}{\sqrt{2}} \cos (ω t)$
$i = \frac{B A ω}{\sqrt{2} R} \sin (ω t)$
$\vec{m} = \frac{B A^{2} ω}{\sqrt{2} R} \sin (ω t) \hat{k}$
$\vec{τ} = \vec{m} \times \vec{B} = \frac{B^{2} A^{2} ω}{\sqrt{2} \times \sqrt{2} R} \sin ω t (\hat{k} \times (\sin ω t \hat{i} + \cos ω t \hat{k}))$
$\vec{τ} = \frac{B^{2} A^{2} ω \sin (ω t)}{2 R} \sin (ω t) \hat{j}$
$\vec{τ} = \frac{B^{2} A^{2} ω}{2 R} \sin^{2} (ω t) \hat{j}$
$\vec{τ} = \frac{α}{4} \hat{j}$
Hence, $III \to s$
Following the same process as in case I for case IV
$ϕ = \frac{B A}{\sqrt{2}} \sin (ω t)$
$i = - \frac{B A ω}{\sqrt{2} R} \cos (ω t)$
$\vec{m} = - \frac{B A^{2} ω}{\sqrt{2} R} \cos ω t (\hat{k})$
$\vec{τ} = \vec{m} \times \vec{B} = - \frac{B^{2} A^{2} ω}{2 R} (\hat{k} \times \hat{j}) \cos^{2} (ω t)$
$\vec{τ} = + \frac{B^{2} A^{2} ω}{2 R} (\hat{i}) \cdot \cos^{2} (\frac{π}{6})$
$\vec{τ} = + \frac{3}{4} α \hat{i}$
Hence, $(IV) \to r$

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	List-I		List-II
(i)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{j} + \cos ω t \hat{k})$	(p)	$0$
(ii)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{i} + \cos ω t \hat{j})$	(q)	$- \frac{α}{4} \hat{i}$
(iii)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{i} + \cos ω t \hat{k})$	(r)	$\frac{3 α}{4} \hat{i}$
(iv)	$\frac{1}{\sqrt{2}} (\cos ω t \hat{j} + \sin ω t \hat{k})$	(s)	$\frac{α}{4} \hat{j}$
		(t)	$- \frac{3 α}{4} \hat{i}$

A small circular loop of area A and resistance R is fixed on a horizontal x y -plane with the center of the…

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