JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A stream of a positively charged particles having \(\frac{ q }{ m }=2 \times 10^{11} \frac{ C }{ kg }\) and velocity \(\overrightarrow{ v }_0=3 \times 10^7 \hat{ i ~ m} / s\) is deflected by an electric field \(1.8 \hat{ j } kV / m\). The electric field exists in a region of \(10 cm\) along \(x\) direction. Due to the electric field, the deflection of the charge particles in the \(y\) direction is \(...........mm\)
- A \(2\)
- B \(4\)
- C \(0.5\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(a =\frac{ F }{ m }=\frac{ qE }{ m }=\left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right)\) \(=3.6 \times 10^{14} m / s ^2\) \(\text { Time to cross plates }=\frac{ d }{ v }\) \(t =\frac{0.10}{3 \times 10^7}\)…
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