Get Answers to all your Questions

header-bg qa

A particle of charge - q & mass m moves in a circle of radius r around an infinitely long line charge of linear charge density + l. Then time period of the revolution of charge will be :

 

Option: 1

T = 2\pi r \sqrt{\frac{m}{2k\lambda q}}


Option: 2

T^{2}= \frac{4 \pi^{2}m}{2k \lambda q}r^{3}


Option: 3

T = \frac{1}{2\pi r}\sqrt{\frac{2k \lambda q}{m}}


Option: 4

T = \frac{1}{2\pi r}\sqrt{\frac{m}{2 \lambda qk}}


Answers (1)

best_answer

From the F.B.D. of charge, we have :

q\left ( \frac{2k \lambda }{r} \right ) =\frac{mv^{2}}{r}

so V = \sqrt{\frac{2kq \lambda}{m}}  Now, T = \frac{2 \pi r }{v}= 2 \pi r\sqrt{\frac{m}{2k \lambda q}}

where k = \frac{1}{4\pi \epsilon_{0}}

Posted by

Rakesh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE