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A charged particle of specific charge (charge/mass) `alpha` released from origin at time `t=0` with velocity `vec v = v_0 (hat i + hat j)` in uniform magnetic field `vec B = B_0 hat i.` Coordinates of the particle at time `t= pi//(B_0 alpha)` are

A

`((v_(0))/(2B_(0)alpha),(sqrt2v_(0))/(alphaB_(0)),(-v)/(B_(0)alpha))`

B

`((-v_(0))/(2B_(0)alpha),0,0,)`

C

`(0,(2v_(0))/(B_(0)alpha),(v_(0)pi)/(2B_(0)alpha))`

D

`((v_(0)pi)/(B_(0)alpha),0,(-2v_(0))/(B_(0)alpha))`

Text Solution

Verified by Experts

The correct Answer is:
D

`alpha=(q)/(m)`, path of the particle will be a helix of time period.

` T=(2pim)/(B_(0)q)=(2pi)/(B_(0)alpha)`
The given time `t=t=(pi)/(B_(0)alpha)=(T)/(2)`
Thus, coordinates of particle at time `t=T//2` would be `(v_(x)T//2,0,-2r)`.
Here `r=(mv_(r))/(B_(0)q)=(v_(0))/(B_(0)alpha)`
Hence, coordinates are `((v_(0))/(B_(0)alpha),0,(-2v)/(B_(0)alpha))`.
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