JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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 \(=9.1 \times 10^{-31} kg\) )
- A \(192\, k\,Vm ^{-1}\)
- B \(192\, m\, Vm ^{-1}\)
- C \(9600\, k\,Vm ^{-1}\)
- D \(16 \,k\,Vm ^{-1}\)
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 \(=\frac{1}{2} mv ^{2}\) \(728\,eV =\frac{1}{2} \times 9.1 \times 10^{-31} \times v ^{2}\)…
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