JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2019
- Hard
- MCQ
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 \(= 9.1\times10^{-31}\, kg\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(12.87\)
Step-by-step Solution
Detailed explanation
\({\text{R}} = \frac{{{\text{mv}}}}{{{\text{qB}}}}\)
\( = \frac{{\sqrt {2{\text{m}}({\text{KE}})} }}{{{\text{qB}}}}\)
\({\text{R}} = \frac{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times \left( {100 \times 1.6 \times {{10}^{ - 18}}} \right)} }}{{1.6 \times {{10}^{ - 10}} \times 1.5 \times {{10}^{ - 3}}}}\)
\({\text{R}} = 2.248\,{\text{cm}}\)
\(\sin \theta = \frac{2}{{2.248}};\,\,\,\,\,\tan \,\theta = \frac{{QU}}{{TU}}\)
\(\frac{2}{{1.026}} = \frac{{QU}}{6}\)
\(QU = 11.69\)
\({\text{PU}} = {\text{R}}(1 - \cos \,\theta )\)
\( = 1.22\)
\({\text{d}} = {\text{QU}} + {\text{PU}}\)
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