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Uniform electric field E and magnetic fi...

Uniform electric field E and magnetic field B, respectively, are along y-axis as shown in the figure. A particle with specific charge q/m leaves the origin O in the direction of x-axis with an initial non-relativistic speed `v_(0)`.

The angle `alpha` between the particle’s velocity vector and y-axis at that moment is:

A

`tan^(-1)((3v_(0)B)/(2pinE))`

B

`tan^(-1)((v_(0)B)/(pi nE))`

C

`tan^(-1)((v_(0)B)/(sqrt(2)pi nE))`

D

`tan^(-1)((v_(0)B)/(2pi nE))`

Text Solution

Verified by Experts

The correct Answer is:
D
As `v_(y)=a_(y)t=((qE)/m)((2pimn)/(qB))=(2pinE)/B`
Thus `tan alpha=(v_(0))/(v_(y))=(v_(0)B)/(2pinE)implies alpha=tan^(-1)((v_(0)B)/(2pin E))`
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