A charged particle of mass m and charge q mov | vedclass

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(B) Using the work-energy theorem: $W_{net} = \Delta K$$qE(2a) = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$$E = \frac{3mv^2}{4qa}$. Thus,

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$(A), (B), (C)$

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(B) Using the work-energy theorem: $W_{net} = \Delta K$$qE(2a) = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$$E = \frac{3mv^2}{4qa}$. Thus,$(A)$ is correct.Rate of work done by electric field $P_E = \vec{F}_E \cdot \vec{v} = (qE\hat{i}) \cdot (v\hat{i}) = qEv = q\left(\frac{3mv^2}{4qa}ight)v = \frac{3mv^3}{4a}$. Thus,$(B)$ is correct.At $Q$,velocity is $-2v\hat{j}$. Electric force is $qE\hat{i}$. Since $\vec{F}_E \perp \vec{v}$,power by electric field is $0$. Magnetic force is always perpendicular to velocity,so power by magnetic field is $0$. Thus,$(C)$ is correct.Angular momentum $L = \vec{r} 	imes \vec{p}$. At $P$,$\vec{r}_P = a\hat{j}$,$\vec{p}_P = mv\hat{i}$,so $L_P = (a\hat{j}) 	imes (mv\hat{i}) = -mav\hat{k}$. Magnitude $|L_P| = mav$.At $Q$,$\vec{r}_Q = 2a\hat{i}$,$\vec{p}_Q = -2mv\hat{j}$,so $L_Q = (2a\hat{i}) 	imes (-2mv\hat{j}) = -4mav\hat{k}$. Magnitude $|L_Q| = 4mav$.Difference $= 4mav - mav = 3mav$. Thus,$(D)$ is incorrect.

List-$I$	List-$II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to current

List-$I$	List-$II$
$(A)$ Permeability of free space	$(I) \ [M L^2 T^{-2}]$
$(B)$ Magnetic field	$(II) \ [M T^{-2} A^{-1}]$
$(C)$ Magnetic moment	$(III) \ [M L T^{-2} A^{-2}]$
$(D)$ Torsional constant	$(IV) \ [L^2 A]$

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