JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Magnetic Effects of Current

A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude

\frac{π}{2} \times 1 0^{- 3}

T. The angle between the direction of magnetic field and velocity of proton is

6 0^{\circ}

. The pitch of the helical path taken by the proton is ____ cm.
(Take, mass of proton

= 1.6 \times 1 0^{- 27}

kg and Charge on proton

= 1.6 \times 1 0^{- 19}

C).

Answer: 40

Solution

From

K E = \frac{1}{2} m v^{2}

v = \frac{2 K E}{m} = \frac{2 \times 2 \times 1.6 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 27}} = 2 \times 1 0^{4}

m s

^{- 1}

. The pitch is the distance advanced along the field in one revolution,

p = (v cos 6 0^{\circ}) T

with

T = \frac{2 π m}{q B}

. Thus

p = 2 \times 1 0^{4} \times \frac{1}{2} \times \frac{2 π \times 1.6 \times 1 0 ^{- 27}}{1.6 \times 1 0 ^{- 19} \times \frac{π}{2} \times 1 0 ^{- 3}} = 0.40

m. The pitch of the helical path is

40

cm.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer