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JEE Mains 26-July-2022 Shift 2 Solved Paper

Section: Physics
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Question : 47 of 90
 
Marks: +1, -0
A velocity selector consists of electric field
→
E
=E
^
k
 and magnetic field
→
B
=B
^
j
 with B=12mT. The value of E required for an electron of energy 728eV moving along the positive x-axis to pass undeflected is :
(Given, mass of electron =9.1×10−31‌kg )
[26-Jul-2022-Shift-2]
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