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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 40.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} K . E = 2 ev, B = \frac{π}{2} \times 10^{- 3} π, θ = 60^{0} \\ \frac{1}{2} {mv}^{2} = 2 ev \\ \frac{1}{2} \times 1.6 \times 10^{- 27} \times v^{2} = 2 \times 1.6 \times 10^{- 19} J \\ v^{2} = 4 \times 10^{8} m / s \\ v = 2 \times 10^{4} m / s \end{array}$
Pitch of the helical path = v_x x T
$= vcos θ \times \frac{2 πm}{Bq} \begin{array}{l} = \frac{2 π \times 1.6 \times 10^{- 27} \times 2 \times 10^{4} \cos 60^{0}}{\frac{π}{2} \times 1.6 \times 10^{- 19}} \\ = 4 \times 10^{- 1} = 0.4 m . \end{array}$
Question Image

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE