Question

An electron is projected at an angle with a uniform magnetic field.If the pitch of the helical path is equal to its radius.Then angle of projection

Answer 1

 

As the particle has entered the magnetic field at angle θ thus we can berak its velocity into two components - vparallel and vperpendicular

 

vparallel = qrB/m

 

Therefore time taken to complete one circle (time period), T = 2pir/v = 2pim/qB

Thus the pitch for the motion can be given by, P = Txvperpendicular

 

P = (2pim/qB)x(vsinθ)

Answer 2

Answer:tan-1(2pi)

Explanation:

We know that :

r=mvsin/qB...................(1)

And... P=Vcos.T...........(2)

Here,

T=2×pi×m/qB............(3)

Now putting equ. (3) in (2),we get that,

P=Vcos×2×pi×m/qB.......(4)

We know,

P=r.........(5)

Now, putting (1)and(4) in (5),we get that,

mVsin/qB=Vcos.2×pi×m/qB

Therefore, sin =cos. 2×pi

Sin/cos=2×pi

tan=2×pi

Theta =tna-1(2pi)

Hence proved

NOTE:use sign of theta every where after sin and cos....... Thanks