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QJEEM20Phca54: JEE Main 2020 Physics - Class 12 JEEM20Phca54 Solution: A charged particle of mass 'm' and charge 'q' moving under the…

A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field Ei→E\overrightarrow i and a uniform magnetic field Bk→B\overrightarrow k follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi→v\overrightarrow i and −2vj→- 2v\overrightarrow j . Then which of the following statements (A, B, C, D) are the correct ?
(Trajectory shown is schematic and not to scale) : JEE Main 2020 (Online) 9th January Morning Slot Physics - Magnetic Effect of Current Question 140 English
(A) E = 34(mv2qa){3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)

(B) Rate of work done by the electric field at P is 34(mv3a){3 \over 4}\left( {{{m{v^3}} \over a}} \right)

(C) Rate of work done by both the fields at Q is zero

(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.

Multiple Choice Options

  • A (A), (B), (C), (D)
  • B (A), (B), (C)
  • C (A), (C), (D)
  • D (B), (C), (D)

Step-by-Step Verified Solution

For the particle moving from point P to point Q,

WElectric field + WMagnetic field = Δ\Delta KE

⇒\Rightarrow qE(2 aa ) + 0 = 12m[(2v)2−v2]{1 \over 2}m\left[ {{{\left( {2v} \right)}^2} - {v^2}} \right]

⇒\Rightarrow 2qE aa = 32mv2{3 \over 2}m{v^2}

⇒\Rightarrow E = 34(mv2qa){3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)

∴\therefore Point (A) is correct.

Rate of work done(Power) by electric

field at p = F→.v→\overrightarrow F .\overrightarrow v = qEv = 34(mv3a){3 \over 4}\left( {{{m{v^3}} \over a}} \right)

∴\therefore Point (B) is also correct.

Angle between electric force and velocity is 90º, hence rate of work done will be zero at Q.

∴\therefore Point (C) is also correct.

Initial angular momentum at P, L→{\overrightarrow L } i = mv a(−k^)a\left( { - \widehat k} \right)

Final angular momentum at Q, L→{\overrightarrow L } f = m(2v) (2a)(−k^)(2a)\left( { - \widehat k} \right)

Change in angular momentum L→{\overrightarrow L } f – L→{\overrightarrow L } i = 3mva(−k^)3mva\left( { - \widehat k} \right)

Magnitude in the change in angular momentum ∣ΔL→∣\left| {\Delta \overrightarrow L } \right| = 3mva3mva

∴\therefore Point (D) is wrong.

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