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Magnetics question

2020 · 6 Sep · Shift 1 · Q40
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Magnetics question

2020 · 6 Sep · Shift 1 · Q40

JEE MainPhysicsMagneticsMCQ+4 / −1
A particle of charge q and mass m is moving with a velocity −vi^- v\widehat i−vi(v eee 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field B→=B0k^\overrightarrow B = {B_0}\widehat kB=B0​k , the maximum value of v for which the particle will not hit the screen is :
  1. A
    2qdB0m{{2qd{B_0}} \over m}m2qdB0​​
  2. B
    qdB03m{{qd{B_0}} \over {3m}}3mqdB0​​
  3. C
    qdB02m{{qd{B_0}} \over {2m}}2mqdB0​​
  4. D
    qdB0m{{qd{B_0}} \over {m}}mqdB0​​
View written solutionFree

Correct answer: D

  1. Set up the motion

The particle starts at some point on the positive xxx-axis and moves toward the screen in the YYY-ZZZ plane, i.e. the plane x=0x=0x=0.

Initial velocity is

v⃗=−vi^\vec v = -v\hat iv=−vi^

and magnetic field is

B⃗=B0k^.\vec B = B_0\hat k.B=B0​k^.

The magnetic force is

F⃗=q(v⃗×B⃗).\vec F = q(\vec v \times \vec B).F=q(v×B).

Now,

(−vi^)×(B0k^)=−vB0(i^×k^).(-v\hat i)\times (B_0\hat k)= -vB_0(\hat i \times \hat k).(−vi^)×(B0​k^)=−vB0​(i^×k^).

Since

i^×k^=−j^,\hat i \times \hat k = -\hat j,i^×k^=−j^​,

we get

v⃗×B⃗=−vB0(−j^)=vB0j^.\vec v \times \vec B = -vB_0(-\hat j)=vB_0\hat j.v×B=−vB0​(−j^​)=vB0​j^​.

So the force is along j^\hat jj^​ (for q>0q>0q>0), always perpendicular to velocity. Hence the particle moves in a circle in the xxx-yyy plane.

  1. Radius of circular path

For motion perpendicular to a uniform magnetic field, radius is

R=mvqB0.R=\frac{mv}{qB_0}.R=qB0​mv​.

(If charge were negative, direction changes but radius remains R=mv∣q∣B0R=\frac{mv}{|q|B_0}R=∣q∣B0​mv​; the options clearly assume positive qqq.)

  1. Geometry of the path

Initially the particle is at distance ddd from the screen, so take initial position as

(x,y)=(d,0).(x,y)=(d,0).(x,y)=(d,0).

Initial velocity is along −i^-\hat i−i^, and force is along +j^+\hat j+j^​, so the center of the circular path lies vertically above the initial point:

center=(d,R).\text{center}=(d,R).center=(d,R).

Thus the trajectory is

(x−d)2+(y−R)2=R2.(x-d)^2+(y-R)^2=R^2.(x−d)2+(y−R)2=R2.

We want the particle to not hit the screen x=0x=0x=0.

The minimum value of xxx on this circle is

xmin⁡=d−R.x_{\min}=d-R.xmin​=d−R.

For the particle to just avoid touching the screen,

xmin⁡=0⇒d−R=0⇒R=d.x_{\min}=0 \quad \Rightarrow \quad d-R=0 \Rightarrow R=d.xmin​=0⇒d−R=0⇒R=d.

So for no collision,

R≤d.R\le d.R≤d.

Using R=mvqB0R=\frac{mv}{qB_0}R=qB0​mv​,

mvqB0≤d.\frac{mv}{qB_0}\le d.qB0​mv​≤d.

Hence the maximum allowed speed is

vmax⁡=qdB0m.v_{\max}=\frac{qdB_0}{m}.vmax​=mqdB0​​.
  1. Match with options
vmax⁡=qdB0mv_{\max}=\frac{qdB_0}{m}vmax​=mqdB0​​

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived answer.

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