JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2022
- Hard
- MCQ
A velocity selector consists of electric field \(\overrightarrow{ E }= E \hat{ k }\) and magnetic field \(\overrightarrow{ B }= B \hat{ j }\) 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, , ass of electron \(=9.1 \times 10^{-31} kg\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(192\, k\,Vm ^{-1}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ E }= E \hat{ k } B =12 mT\)
\(\overrightarrow{ B }= B \hat{ j } \text { Energy }=728\,eV\)
Energy \(=\frac{1}{2} mv ^{2}\)
\(728\,eV =\frac{1}{2} \times 9.1 \times 10^{-31} \times v ^{2}\)
\(728 \times 1.6 \times 10^{-19}=\frac{1}{2} \times 9.1 \times 10^{-31} \times v ^{2}\)
\(v =16 \times 10^{6} m / s\)
\(E=v B\)
\(E =16 \times 10^{6} \times 12 \times 10^{-3}\)
\(E =192 \times 10^{3}\,V / m\)
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