Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph627d: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph627d Solution: A small circular loop of area and resistance is fixed on a…

A small circular loop of area AA and resistance RR is fixed on a horizontal xyx y -plane with the center of the loop always on the axis n^\hat{n} of a long solenoid. The solenoid has mm turns per unit length and carries current II counterclockwise as shown in the figure. The magnetic field due to the solenoid is in n^\hat{n} direction. List-I gives time dependences of n^\hat{n} in terms of a constant angular frequency ω\omega . List-II gives the torques experienced by the circular loop at time t=π6ωt=\frac{\pi}{6 \omega} . Let α=A2μ02m2I2ω2R\alpha=\frac{A^{2} \mu_{0}^{2} m^{2} I^{2} \omega}{2 R} .

JEE Advanced 2022 Paper 1 Online Physics - Magnetism Question 11 English

List-I List-II
(I) 12(sin⁡ωtȷ^+cos⁡ωtk^)\frac{1}{\sqrt{2}}(\sin \omega t \hat{\jmath}+\cos \omega t \hat{k}) (P) 0
(II) 12(sin⁡ωtı^+cos⁡ωtȷ^)\frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{\jmath}) (Q) −α4ı^-\frac{\alpha}{4} \hat{\imath}
(III) 12(sin⁡ωtı^+cos⁡ωtk^)\frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{k}) (R) 3α4ı^\frac{3 \alpha}{4} \hat{\imath}
(IV) 12(cos⁡ωtȷ^+sin⁡ωtk^)\frac{1}{\sqrt{2}}(\cos \omega t \hat{\jmath}+\sin \omega t \hat{k}) (S) α4ȷ^\frac{\alpha}{4} \hat{\jmath}
(T) −3α4ı^-\frac{3 \alpha}{4} \hat{\imath}

Which one of the following options is correct?

Multiple Choice Options

  • A I →\rightarrow Q, II →\rightarrow P, III →\rightarrow S, IV →\rightarrow T
  • B I→S,II→T\mathrm{I} \rightarrow \mathrm{S}, \mathrm{II} \rightarrow \mathrm{T} , III →Q\rightarrow \mathrm{Q} , IV →P\rightarrow \mathrm{P}
  • C I→Q,II→P\mathrm{I} \rightarrow \mathrm{Q}, \mathrm{II} \rightarrow \mathrm{P} , III →S\rightarrow \mathrm{S} , IV →R\rightarrow \mathrm{R}
  • D I→T\mathrm{I} \rightarrow \mathrm{T} , II →Q\rightarrow \mathrm{Q} , III →P\rightarrow \mathrm{P} , IV →R\rightarrow \mathrm{R}

Step-by-Step Verified Solution

I. ϕ=B A k^ . n^\phi = B\,A\,\widehat k\,.\,\widehat n

ϕ=BA2cos⁡(ωt)\phi = {{BA} \over {\sqrt 2 }}\cos (\omega t)

ε=BAω2sin⁡(ωt)\varepsilon = {{BA\omega } \over {\sqrt 2 }}\sin (\omega t)

i=BAω2Rsin⁡(ωt)i = {{BA\omega } \over {\sqrt 2 R}}\sin (\omega t)

m→=iA k^=BA2ω2Rsin⁡(ωt)k^\overrightarrow m = iA\,\widehat k = {{B{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)\widehat k

τ→=m→×B→=B2A2ω2Rsin⁡(ωt)(k^×n^)\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = {{{B^2}{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)(\widehat k \times \widehat n)

=−B2A2ω2R[i^]sin⁡2(ωt)= - {{{B^2}{A^2}\omega } \over {2R}}\left[ {\widehat i} \right]{\sin ^2}(\omega t)

τ=−B2A2ω2R[sin⁡2(π6)]=−α4i^\tau = - {{{B^2}{A^2}\omega } \over {2R}}\left[ {{{\sin }^2}\left( {{\pi \over 6}} \right)} \right] = {{ - \alpha } \over 4}\widehat i

(I) →\to Q.

II. ϕ=O\phi = O

(II) →\to P.

III. ϕ=BAω2cos⁡(ωt)\phi = {{BA\omega } \over {\sqrt 2 }}\cos (\omega t)

i=BAω2Rsin⁡(ωt)i = {{BA\omega } \over {\sqrt 2 R}}\sin (\omega t)

m→=BA2ω2Rsin⁡(ωt)k^\overrightarrow m = {{B{A^2}\omega } \over {\sqrt 2 R}}\sin (\omega t)\widehat k

τ→=m→×B→=B2A2ω2×2Rsin⁡ωt(k^×(sin⁡ωti^+cos⁡ωtk^))\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = {{{B^2}{A^2}\omega } \over {\sqrt 2 \times \sqrt 2 R}}\sin \omega t(\widehat k \times (\sin \omega t\widehat i + \cos \omega t\widehat k))

τ=B2A2ωsin⁡(ωt)2Rsin⁡(ωt)j^\tau = {{{B^2}{A^2}\omega \sin (\omega t)} \over {2R}}\sin (\omega t)\widehat j

=B2A2ω2Rsin⁡2(ωt)j^= {{{B^2}{A^2}\omega } \over {2R}}{\sin ^2}(\omega t)\widehat j

=α4j^= {\alpha \over 4}\widehat j

(III) →\to S.

IV. ϕ=BA2sin⁡(ωt)\phi = {{BA} \over {\sqrt 2 }}\sin (\omega t)

i=−BAω2Rcos⁡(ωt)i = - {{BA\omega } \over {\sqrt 2 R}}\cos (\omega t)

m→=−BA2ω2Rcos⁡ωt(k^)\overrightarrow m = - {{B{A^2}\omega } \over {\sqrt 2 R}}\cos \omega t(\widehat k)

τ→=m→×B→=−B2A2ω2R(k^×j^)cos⁡2(ωt)\overrightarrow \tau = \overrightarrow m \times \overrightarrow B = - {{{B^2}{A^2}\omega } \over {2R}}(\widehat k \times \widehat j){\cos ^2}(\omega t)

τ=+B2A2ω2R(i^) . cos⁡2(π6)\tau = + {{{B^2}{A^2}\omega } \over {2R}}(\widehat i)\,.\,{\cos ^2}\left( {{\pi \over 6}} \right)

=+34αi^= + {3 \over 4}\alpha \widehat i

(IV) →\to R.

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