Question 288: Magnetism and Matter

Question

A proton with a kinetic energy of $$2.0 ~\mathrm{eV}$$ moves into a region of uniform magnetic field of magnitude $$\frac{\pi}{2} 	imes 10^{-3} \mathrm{~T}$$. The angle between the direction of magnetic field and velocity of proton is $$60^{\circ}$$. The pitch of the helical path taken by the proton is __________ $$\mathrm{cm}$$. (Take, mass of proton $$=1.6 	imes 10^{-27} \mathrm{~kg}$$ and Charge on proton $$=1.6 	imes 10^{-19} \mathrm{C}$$ ).

Accepted Answer

Given a proton with a kinetic energy of 2 eV, moving into a region of uniform magnetic field of magnitude $$\frac{\pi}{2} 	imes 10^{-3} T$$, and with an angle of $$60^{\circ}$$ between the direction of the magnetic field and the velocity of the proton, we want to determine the pitch of the helical path taken by the proton.
First, calculate the proton's speed (v) using the kinetic energy (K.E) formula:
$$v = \sqrt{\frac{2 	imes KE}{m}}$$
Next, find the component of the velocity in the direction of the magnetic field (parallel component):
$$v_{\parallel} = v \cos 	heta$$
In this case, ÃÂ¸ is given as $$60^{\circ}$$, so $$\cos 	heta = \frac{1}{2}$$.
The pitch of a charged particle moving in a magnetic field with an angle ÃÂ¸ to the direction of the magnetic field is given by the formula:
$$p = \frac{2 \pi m v_{\parallel}}{qB}$$
Substitute the values for the mass of the proton (m), the kinetic energy (KE), the charge of the proton (q), and the magnetic field (B) into the formula:
$$p = \frac{2 \pi 	imes \sqrt{2mKE} 	imes \frac{1}{2} 	imes 2}{qB}$$
After substituting the given values, the pitch of the helical path is found to be:
$$p = 0.4 \, m = 40 \, cm$$
In conclusion, the pitch of the helical path taken by the proton in the magnetic field is 40 cm.

Magnetism and Matter

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