Verified Solution • 2025–26 Session moderate

QJEEM24Phd42a: JEE Main 2024 Physics - Class 12 JEEM24Phd42a Solution: A proton moving with a constant velocity passes through a…

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E→\overrightarrow{\mathrm{E}} and B→\overrightarrow{\mathrm{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}=0

(B) E=0, B≠0\mathrm{E}=0, \mathrm{~B} \neq 0

(C) E≠0, B=0\mathrm{E} \neq 0, \mathrm{~B}=0

(D) E≠0, B≠0\mathrm{E} \neq 0, \mathrm{~B} \neq 0

Choose the most appropriate answer from the options given below :

Multiple Choice Options

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

Step-by-Step Verified Solution

Net force on particle must be zero i.e. qE→+qV→×B→=0\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0

Possible cases are

(i) E→&B→=0\overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}=0

(ii) V→×B→=0,E→=0\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0, \overrightarrow{\mathrm{E}}=0

(iii) qE→=−qV→×B→q \overrightarrow{\mathrm{E}}=-\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}

E→≠0&B→≠0\overrightarrow{\mathrm{E}} \neq 0 \& \overrightarrow{\mathrm{B}} \neq 0

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