Question

A beam of protons with a velocity of $4\times {10}^{5}m/s$ enters in a region of uniform magnetic field of 0.3 T. The velocity makes an angle of ${60}^0$ with the magnetic field. Find the radius of the helical path taken by the proton beam and the pitch of the helix

• A. 2.2 cm and 4.4 cm respectively
• B. 1.3 cm and 6.4 cm respectively
• C. 1.5 cm and 5.4 cm respectively
• D. 1.2 cm and 4.4 cm respectively

Answer 1

The correct option is D 1.2 cm and 4.4 cm respectively


Radius of helical path: $R=\frac{m{v}_{\perp }}{qB}$
$v_{\perp }=vsin{60}^0$$=4\times {10}^5\times \frac{\sqrt{3}}{2}$
$v_{\perp }=2\sqrt{3}\times {10}^{5}m/s$

$v_{\parallel }=vcos{60}^0$$=4\times {10}^5\times \frac{1}{2}$
$v_{\parallel }=2\times {10}^{5}m/s$

m=$1.6\times {10}^{-27}kg$
$v_{\perp }=2\sqrt{3}\times {10}^{5}m/s$
q =$1.6\times {10}^{-19}C$
B=0.3T
R=$\frac{(1.6\times {10}^{-27})\times (2\sqrt{3}\times {10}^5)}{1.6\times {10}^{-19}\times 0.3}$
R=1.2cm $\to$ Radius of the helical path

$T$=Time period of revolution
$T=\frac{2\pi r}{v_{\perp }}=\frac{2\times 3.14\times 0.012}{2\sqrt{3}\times {10}^5}$
Pitch =$v_{\parallel }\times T$
$=\frac{2\times {10}^5\times 2\times 3.14\times 0.012}{2\sqrt{3}\times {10}^5}$
Pitch = 4.4cm