Magnetic Effects & Magnetism

JEE Main level, single correct, medium

156 of 276

Question 156 of 276

A charged particle of mass ‘ mm ’ and charge ‘ qq ’ moving under the influence of uniform electric field E⃗i^\vec{\text{E}}\hat{i} and a uniform magnetic field B⃗k^\vec{\text{B}}\hat{k} follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi⃗v\vec{i} and −2vj⃗-2v\vec{j} . Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale)



(A)\left(A\right) E=32(mv2qa)E=\dfrac{3}{2}\left(\dfrac{m{v}^{2}}{qa}\right)
(B)\left(B\right) Rate of work done by the electric field at P is 32(mv3a)\dfrac{3}{2}\left(\dfrac{m{v}^{3}}{a}\right)
(C)\left(C\right) Rate of work done by both the fields at Q is zero
(D)\left(D\right) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav2mav .

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Answer: (C) (A),(B),(C)\left(A\right),\left(B\right),\left(C\right)

(A)(A) by work energy theorem
Wmag+Wele=12m(2v)2−12m(v)2{W}_{mag}+{W}_{ele}=\dfrac{1}{2}m(2v{)}^{2}-\dfrac{1}{2}m(v{)}^{2}
0+qE02a=32mv20+{qE}_{0}2a=\dfrac{3}{2}{mv}^{2}
E0=34mv2qa{E}_{0}=\dfrac{3}{4}\dfrac{{mv}^{2}}{qa}
(B)(B) Rate of work done at A=A= power of electric force
=qE0V={qE}_{0}V
=34mv3a=\dfrac{3}{4}\dfrac{{mv}^{3}}{a}
(C)(C) at Q,dwdt=0Q,\dfrac{dw}{dt}=0 for both forces
(D)\left(D\right) ΔL⃗=(−m2v2ak^)−(−mvak)\Delta \vec{L}=(-m2\text{v}2a\hat{k})-(-\text{mvak})
∣ΔL⃗∣=3mva|\Delta \vec{L}|=3mva

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