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Physics Question on Magnetism and matter

A proton is projected with speed v in magnetic field B of magnitude 1 T. The angle between velocity and magnetic field is 600 as shown below. The kinetic energy of a proton is 2 eV (mass of proton = 1.67×10−27kg1.67\times10^{-27} kg, e = 1.6×10−191.6\times10^{-19}C). The pitch of the path of the proton is approximately?

A

6.28 x 10-2 m

B

6.28 x 10-4 m

C

3.14 x 10-2 m

D

3.14 x 10-4 m

Answer

6.28 x 10-4 m

Explanation

Solution

The correct option is (B) : 6.28×10−46.28\times10^{-4} m
R=mv sin60∘qBR = \frac{mv\, sin60^{\circ}}{qB}
T−2πmqB     K.E=−12mv2T-\frac{2\pi m}{qB}\,\,\,\,\,K.E = -\frac{1}{2}mv^2
pitch = v cosθ×(2πmqB)        2KM=vv\,cos\theta\times(\frac{2\pi m}{qB})\,\,\,\,\,\,\,\,\sqrt{\frac{2K}{M}}=v
2KM×12×2πmqB\sqrt{\frac{2K}{M}}\times\frac{1}{2}\times\frac{2\pi m}{qB}
= πeB.2Km=3.141.6×10−19×1×2×2×(1.6)2×10−46\frac{\pi}{eB}.\sqrt{2Km}=\frac{3.14}{1.6\times10^{-19}\times1}\times\sqrt{2\times2\times(1.6)^2\times10^{-46}}
=3.141.6×10−19×2×1.6×10−23=\frac{3.14}{1.6\times10^{-19}}\times2\times1.6\times10^{-23}
=6.28×10−4m=6.28\times10^{-4}m