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Magnetics question

2020 · 9 Jan · Shift 1 · Q52
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  5. /2020 · 9 Jan · Shift 1 · Q52

Magnetics question

2020 · 9 Jan · Shift 1 · Q52

JEE MainPhysicsMagneticsMCQ+4 / −1
A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field Ei→E\overrightarrow iEi and a uniform magnetic field Bk→B\overrightarrow kBk follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi→v\overrightarrow ivi and −2vj→- 2v\overrightarrow j−2vj​. 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 150 English (A) E =34(mv2qa){3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)43​(qamv2​)(B) Rate of work done by the electric field at P is 34(mv3a){3 \over 4}\left( {{{m{v^3}} \over a}} \right)43​(amv3​) (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.
  1. A
    (A), (B), (C), (D)
  2. B
    (A), (B), (C)
  3. C
    (A), (C), (D)
  4. D
    (B), (C), (D)
View written solutionFree

Correct answer: B

  1. Given data
  • Uniform electric field: E⃗=E i^\vec E = E\,\hat iE=Ei^
  • Uniform magnetic field: B⃗=B k^\vec B = B\,\hat kB=Bk^
  • At point PPP: v⃗P=v i^\vec v_P = v\,\hat ivP​=vi^
  • At point QQQ: v⃗Q=−2v j^\vec v_Q = -2v\,\hat jvQ​=−2vj^​
  • From the figure, let the horizontal separation between the two vertical lines through PPP and QQQ be aaa.

We use:

F⃗=q(E⃗+v⃗×B⃗)\vec F = q(\vec E + \vec v \times \vec B)F=q(E+v×B)

Since magnetic force does no work, only electric field changes kinetic energy.

  1. Use work-energy theorem to find EEE

At PPP, speed =v=v=v.

At QQQ, speed =2v=2v=2v.

So change in kinetic energy is

ΔK=12m(2v)2−12mv2=12m(4v2−v2)=32mv2\Delta K = \frac12 m(2v)^2 - \frac12 mv^2 = \frac12 m(4v^2-v^2) = \frac32 mv^2ΔK=21​m(2v)2−21​mv2=21​m(4v2−v2)=23​mv2

Only electric field does work:

WE=qE ΔxW_E = qE\,\Delta xWE​=qEΔx

From the figure, displacement along xxx from PPP to QQQ is aaa. Hence

qEa=32mv2qEa = \frac32 mv^2qEa=23​mv2

So

E=32mv2qaE = \frac{3}{2}\frac{mv^2}{qa}E=23​qamv2​

This does not match statement (A), which says

E=34mv2qaE=\frac34\frac{mv^2}{qa}E=43​qamv2​

Therefore, (A) is false.

  1. Check statement (B): rate of work done by electric field at PPP

Power delivered by electric field is

PE=F⃗E⋅v⃗=qE⃗⋅v⃗P_E = \vec F_E\cdot \vec v = q\vec E\cdot \vec vPE​=FE​⋅v=qE⋅v

At PPP:

v⃗P=vi^,E⃗=Ei^\vec v_P = v\hat i, \qquad \vec E = E\hat ivP​=vi^,E=Ei^

Thus

PP=qEvP_P = qEvPP​=qEv

Using the value of EEE obtained from energy relation:

PP=qv(32mv2qa)=32mv3aP_P = qv\left(\frac{3}{2}\frac{mv^2}{qa}\right) = \frac{3}{2}\frac{mv^3}{a}PP​=qv(23​qamv2​)=23​amv3​

This does not match statement (B), which says

34mv3a\frac34\frac{mv^3}{a}43​amv3​

So (B) is false by direct calculation.

  1. Check statement (C): rate of work done by both fields at QQQ

At QQQ, velocity is

v⃗Q=−2vj^\vec v_Q = -2v\hat jvQ​=−2vj^​

Power due to electric field:

PE=qE⃗⋅v⃗Q=q(Ei^)⋅(−2vj^)=0P_E = q\vec E\cdot \vec v_Q = q(E\hat i)\cdot(-2v\hat j)=0PE​=qE⋅vQ​=q(Ei^)⋅(−2vj^​)=0

Power due to magnetic field is always zero because

q(v⃗×B⃗)⋅v⃗=0q(\vec v\times \vec B)\cdot \vec v = 0q(v×B)⋅v=0

Hence total rate of work done by both fields at QQQ is

0+0=00+0=00+0=0

Therefore, (C) is true.

  1. Check statement (D): difference in magnitude of angular momentum

Angular momentum about the origin shown in the figure depends on position. From the usual figure for this problem, point PPP lies on the yyy-axis at height aaa, and point QQQ lies on the xxx-axis at distance aaa from the origin.

Then:

At PPP:

r⃗P=aj^,p⃗P=mvi^\vec r_P = a\hat j, \qquad \vec p_P = mv\hat irP​=aj^​,p​P​=mvi^

So

L⃗P=r⃗P×p⃗P=aj^×mvi^=−mavk^\vec L_P = \vec r_P\times \vec p_P = a\hat j \times mv\hat i = -mav\hat kLP​=rP​×p​P​=aj^​×mvi^=−mavk^

Hence

∣LP∣=mav|L_P| = mav∣LP​∣=mav

At QQQ:

r⃗Q=ai^,p⃗Q=−2mvj^\vec r_Q = a\hat i, \qquad \vec p_Q = -2mv\hat jrQ​=ai^,p​Q​=−2mvj^​

So

L⃗Q=ai^×(−2mvj^)=−2mavk^\vec L_Q = a\hat i \times (-2mv\hat j) = -2mav\hat kLQ​=ai^×(−2mvj^​)=−2mavk^

Hence

∣LQ∣=2mav|L_Q| = 2mav∣LQ​∣=2mav

Therefore difference in magnitudes is

∣∣LQ∣−∣LP∣∣=2mav−mav=mav\big||L_Q|-|L_P|\big| = 2mav-mav = mav​∣LQ​∣−∣LP​∣​=2mav−mav=mav

not 2mav2mav2mav.

So (D) is false.

  1. Final evaluation of options
  • (A) False
  • (B) False
  • (C) True
  • (D) False

Thus none of the listed options matches exactly, since every option includes extra false statements.

The only definitely correct statement is (C).

  1. Comparison with stored answer

Stored correct answer is B, i.e. (A),(B),(C)(A),(B),(C)(A),(B),(C).

From the physics calculations above, I do not agree with that stored answer. Statements (A) and (B) are inconsistent with work-energy theorem unless some figure detail is very different from the standard interpretation of aaa.

So the stored answer appears incorrect or the figure/parameter aaa is missing crucial geometric information.

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