Skip to main content
Back to MAGNETIC EFFECT OF CURRENT

JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

1 Medium
1 of 1
Application +4 | -1
A particle of charge qq and mass mm starts moving from the origin under the action of an electric field, E=E0i^E=E_0 \hat{{i}} and B=B0i^B=B_0 \hat{{i}} with a velocity, v=v0j^v=v_0 \hat{{j}}. The speed of the particle will become 52v0\frac{\sqrt{5}}{2} v_0 after a time
View answer and explanation
Correct option: B
mv02qE\frac{m v_0}{2 q E}
Option B (mv02qE\frac{m v_0}{2 q E}) is correct because it is derived from calculating the time when the particle's speed reaches the given value under the influence of electric and magnetic fields.

More Questions from MAGNETIC EFFECT OF CURRENT

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.