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Magnetics question

2024 · 27 Jan · Shift 1 · Q64
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  5. /2024 · 27 Jan · Shift 1 · Q64

Magnetics question

2024 · 27 Jan · Shift 1 · Q64

JEE MainPhysicsMagneticsMCQ+4 / −1
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E→\overrightarrow{\mathrm{E}}E and B→\overrightarrow{\mathrm{B}}B represent the electric and magnetic fields respectively, then the region of space may have : (A) E=0, B=0\mathrm{E}=0, \mathrm{~B}=0E=0, B=0(B) E=0, Beq0\mathrm{E}=0, \mathrm{~B} eq 0E=0, Beq0(C) Eeq0, B=0\mathrm{E} eq 0, \mathrm{~B}=0Eeq0, B=0(D) Eeq0, Beq0\mathrm{E} eq 0, \mathrm{~B} eq 0Eeq0, Beq0 Choose the most appropriate answer from the options given below :
  1. A
    (A), (B) and (C) only
  2. B
    (A), (C) and (D) only
  3. C
    (A), (B) and (D) only
  4. D
    (B), (C) and (D) only
View written solutionFree

Correct answer: C

  1. Condition for no change in velocity

For a charged particle moving in electric and magnetic fields, the Lorentz force is

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

If the proton passes through the region without any change in its velocity, then its acceleration is zero, so the net force must be zero:

E⃗+v⃗×B⃗=0.\vec{E} + \vec{v} \times \vec{B} = 0.E+v×B=0.

So we check each case.


  1. Case (A): E=0,  B=0E=0,\; B=0E=0,B=0

Then

F⃗=q(0+v⃗×0)=0.\vec{F} = q(0 + \vec{v} \times 0)=0.F=q(0+v×0)=0.

So the proton can move with constant velocity.

✅ Possible


  1. Case (B): E=0,  B≠0E=0,\; B\neq 0E=0,B=0

Then

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

This can be zero if

v⃗∥B⃗\vec{v} \parallel \vec{B}v∥B

(or antiparallel), because then

v⃗×B⃗=0.\vec{v} \times \vec{B} = 0.v×B=0.

Hence constant velocity is possible.

✅ Possible


  1. Case (C): E≠0,  B=0E\neq 0,\; B=0E=0,B=0

Then

F⃗=qE⃗.\vec{F} = q\vec{E}.F=qE.

Since E⃗≠0\vec{E} \neq 0E=0, the force is nonzero, so the proton must accelerate and its velocity will change.

❌ Not possible


  1. Case (D): E≠0,  B≠0E\neq 0,\; B\neq 0E=0,B=0

Then it is possible that

E⃗=−v⃗×B⃗,\vec{E} = -\vec{v} \times \vec{B},E=−v×B,

so that the net force becomes zero:

q(E⃗+v⃗×B⃗)=0.q(\vec{E} + \vec{v} \times \vec{B}) = 0.q(E+v×B)=0.

Thus the proton can still move with unchanged velocity.

✅ Possible


  1. Final selection

Possible cases are:

  • (A)
  • (B)
  • (D)

So the correct option is

C: (A), (B) and (D) only.\boxed{\text{C: (A), (B) and (D) only}}.C: (A), (B) and (D) only​.
  1. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.

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