JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
An electron, moving along the \(x-\) axis with an initial energy of \(100\, eV\), enters a region of magnetic field \(\vec B = (1.5\times10^{-3}T)\hat k\) at \(S\) (See figure). The field extends between \(x = 0\) and \(x = 2\, cm\). The electron is detected at the point \(Q\) on a screen placed \(8\, cm\) away from the point \(S\). The distance \(d\) between \(P\) and \(Q\) (on the screen) is :......\(cm\) (electron's charge \(= 1.6\times10^{-19}\, C\), mass of electron \(= 9.1\times10^{-31}\, kg\))

- A \(1.22\)
- B \(2.25\)
- C \(12.87\)
- D \(11.65\)
Answer & Solution
Correct Answer
(C) \(12.87\)
Step-by-step Solution
Detailed explanation
\({\text{R}} = \frac{{{\text{mv}}}}{{{\text{qB}}}}\) \( = \frac{{\sqrt {2{\text{m}}({\text{KE}})} }}{{{\text{qB}}}}\)…
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