JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
A particle of charge '-q' and mass 'm' moves in a circle of radius 'r' around an infinitely long line charge of linear charge density '+λ'. Then time period will be given as :
(Consider k as Coulomb's constant)
- (a) T²=(4 π² m)/(2 k λ q) r³
- (b) T=1/(2 π r) √(m/(2 k λ q))
- (c) T=1/(2 π) √((2 k λ q)/m)
- (d) T=2 π r √(m/(2 k λ q))
Answer and solution
Answer (d) T=2 π r √(m/(2 k λ q))
Solution
Physics sir · 3 steps
Let the line charge's pull supply the centripetal force; the speed comes out the same at every radius.
- Field of the line. By Gauss's law, an infinite line charge gives a radial field E = λ/(2πε₀ r) = 2kλ/r. The charge -q is pulled towards the line with force (2kλ q)/r.
- Force balance. This attraction is the centripetal force, so mv²/r = (2kλ q)/r. The r cancels, leaving v = √((2kλ q)/m), which does not depend on the radius.
T = (2π r)/v = 2π r√(m/(2kλ q))
- So. The period grows in direct proportion to r, which matches option (d).
The idea to keep. A line charge's field falls as 1/r, not 1/r². So orbits around it have the same speed at every radius, and the period grows in direct proportion to r.
Answer from the NTA final answer key. Solution by PrepWiser.