QJEEM23Ph2727: JEE Main 2023 Physics - Class 12 JEEM23Ph2727 Solution: A proton with a kinetic energy of 2.0   eV moves into a…

Given a proton with a kinetic energy of 2 eV, moving into a region of uniform magnetic field of magnitude $$\frac{\pi }{2}\times {10}^{-3}T$$, 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.

1. First, calculate the proton's speed (v) using the kinetic energy (K.E) formula:

$$v=\sqrt{\frac{2\times KE}{m}}$$

1. Next, find the component of the velocity in the direction of the magnetic field (parallel component):

$$v_{\parallel }=v\cos \theta$$

In this case, θ is given as $${60}^{\circ }$$, so $$\cos \theta =\frac{1}{2}$$.

1. The pitch of a charged particle moving in a magnetic field with an angle θ to the direction of the magnetic field is given by the formula:

$$p=\frac{2\pi m{v}_{\parallel }}{qB}$$

1. 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:

$$p=\frac{2\pi \times \sqrt{2mKE}\times \frac{1}{2}\times 2}{qB}$$

1. After substituting the given values, the pitch of the helical path is found to be:

$$p=0.4m=40cm$$

In conclusion, the pitch of the helical path taken by the proton in the magnetic field is 40 cm.