wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A velocity selector consists of electric field E=E^k and magnetic field B=B^j with B=12 mT. The value of E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is (Given, mass of electron =9.1×1031 kg)

A
9600 kVm1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
192 mVm1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
192 kVm1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
16 kVm1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 192 kVm1
As we know that,

velocity of selector, v=EB and

Kinetic energy, K=12mV2

2Km×B=E

E=2×728×1.6×10199.1×1031×12×103

=192000 V/m

Hence, (A) is the correct answer.

flag
Suggest Corrections
thumbs-up
22
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Magnetic Force
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon