Correct Answer is Option (A)

When a charged particle moves in a magnetic field, its motion is affected by the components of its velocity that are parallel and perpendicular to the magnetic field.

The component of velocity that is parallel to the magnetic field doesn't get affected by the magnetic field. It causes the particle to move along the magnetic field lines in a straight line.

The component of velocity that is perpendicular to the magnetic field causes the charged particle to move in a circular path around the magnetic field lines due to the magnetic Lorentz force.

Combining these two effects, the charged particle follows a helical path, where the axis of the helix is aligned with the magnetic field.

Correct Answer is 40

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.

Correct Answer is 144

$r=\frac{mv}{qB}=\frac{\sqrt{2km}}{qB},$

$m=\frac{r^2{q}^2{B}^2}{2k}$



$\begin{array}{rl}\mathrm{m}=& \frac{\frac{1}{100}\times \frac{3}{100}\times 2\times 2\times 4\times {10}^{-3}\times 4\times {10}^{-3}\times {10}^{-12}}{2\times (100)\times {10}^{-6}}\\ \\ & =144\times {10}^{-18}\mathrm{kg}\end{array}$

Correct Answer is 10

$R=\frac{mv}{qB}$

$R=\frac{\sqrt{2mKE}}{qB}$

$=\frac{\sqrt{2\times 24\times 1.67\times {10}^{-27}\times 5\times 1.6\times {10}^{-16}}}{1.6\times {10}^{-19}\times 0.5}$

= 10.009 cm = 10 cm

Correct Answer is 2

$R=\frac{\sqrt{2mK}}{qB}$

So, $\frac{r_d}{r_p}=\frac{\sqrt{m_d}/q_d}{\sqrt{m_p}/q_p}$

$=\sqrt{2}$

So $x=2$