JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Medium
- Numerical
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 \(...........mm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(a =\frac{ F }{ m }=\frac{ qE }{ m }=\left(2 \times 10^{11}\right)\left(1.8 \times 10^3\right)\)
\(=3.6 \times 10^{14} m / s ^2\)
\(\text { Time to cross plates }=\frac{ d }{ v }\)
\(t =\frac{0.10}{3 \times 10^7}\)
\(y =\frac{1}{2} at ^2=\frac{1}{2}\left(3.6 \times 10^{14}\right)\left(\frac{0.01}{9 \times 10^{14}}\right)\)
\(=0.2 \times 0.01\)
\(=0.002\,m\)
\(=2\,mm\)
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