Q.

A velocity selector consists of electric field E=Ek^ and magnetic field B=Bj^ with B = 12 mT. The value E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is :

(Given, mass of electron = 9.1×10–31 kg)

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

a

192 kVm–1

b

16 kVm–1

c

9600 kVm–1

d

192 m Vm–1

answer is A.

(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

E=Ek^ B=12mTB=Bj^  Energy =728eV Energy =12mv2728eV=12×9.1×1031×v2728×1.6×1019=12×9.1×1031×v2v=16×106m/sE=vBE=16×106×12×103E=192×103V/m

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon