Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

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

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 40.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

detailed_solution_thumbnail

$\begin{array}{l} K . E = 2 ev, B = \frac{π}{2} \times 10^{- 3} π, θ = 60^{0} \\ \frac{1}{2} {mv}^{2} = 2 ev \\ \frac{1}{2} \times 1.6 \times 10^{- 27} \times v^{2} = 2 \times 1.6 \times 10^{- 19} J \\ v^{2} = 4 \times 10^{8} m / s \\ v = 2 \times 10^{4} m / s \end{array}$
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

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!

A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π2×10−3T. The angle between the direction of magnetic field and velocity of proton is 60o.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 ).