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In the previous problem, if time period,...

In the previous problem, if time period, `T =2(pi) m/(BQ)` where Q is the charge of the particle and m is its mass, the ratio of time spent by the particle in field when `(theta)` is positive to when `(theta)` is negative is given by

A

`((pi//2+theta)/(pi//2-theta))`

B

`((pi+theta)/(pi-theta))`

C

`((pi-theta)/(pi+theta))`

D

`((pi//2-theta)/(pi//2+theta))`

Text Solution

Verified by Experts

The correct Answer is:
A
When `theta "is" +ve, "angle at centre" =pi+2 theta`
When `theta "is" +ve, "angle at centre" =pi-2 theta`
When `theta = 0 ` incidence is normal a semicircle is complete
and `T=2 pi (m)/(BQ) ("time period")`
when `theta` is positive time spent in magnetic field
`(T)/(2 pi)xx(pi +2 theta)`
When `theta` is negative time spent in magnetic field
`=(T)/(2 pi)xx(pi - 2 theta)`
Hence, ratio`=(pi+2 theta)/(pi - 2 theta)=(pi//2+theta)/(pi//2-theta)`.
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