Magnetics Ques 36
- A beam of protons with a velocity $4 \times 10^{5}$ $m / s$ enters a uniform magnetic field of $0.3$ $T$ at an angle of $60^{\circ}$ to the magnetic field. Find the radius of the helical path taken by the protons beam. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation).
(1986, 6M)
Show Answer
Answer:
Correct Answer: 36.$ 36.2 \times 10^{-2} m , \quad 4.37 \times 10^{-2} m $
Solution:
Formula:
Magnetic Force Acting On A Moving Point Charge:
- (i) $r=\frac{m v \sin \theta}{B q}=\frac{\left(1.67 \times 10^{-27}\right)\left(4 \times 10^{5}\right)\left(\sin 60^{\circ}\right)}{(0.3)\left(1.6 \times 10^{-19}\right)}$
$ =1.2 \times 10^{-2} m $
(ii) $p=(\frac{2 \pi m}{B q})(v \cos \theta)$
$ \begin{aligned} & =\frac{(2 \pi)\left(1.67 \times 10^{-27}\right)\left(4 \times 10^{5}\right)\left(\cos 60^{\circ}\right)}{(0.3)\left(1.6 \times 10^{-19}\right)} \\ & =4.37 \times 10^{-2} m \end{aligned} $
BETA
