\(y=0\) અને \(y = d\) વચ્ચેનો વિસ્તાર ચુંબકીય ક્ષેત્ર \(\vec B = B\hat z\) ધરાવે છે. \(m\) દળ અને \(q\) વિજભાર ધરાવતો એક કણ \(\vec v = v\hat i\) વેગથી આ વિસ્તારમાં પ્રવેશે છે. જો \(d = \frac{{mv}}{{2qB}}\) , હોય તો આ વિસ્તારની બીજી બાજુએ નિર્ગમન બિંદુએ વિજભારીત કણનો પ્રવેગ કેટલો હશે?

A \(\frac{{qvB}}{m}\,\left( {\,\frac{{ \hat j + \hat i}}{{\sqrt 2 }}} \right)\)
B \(\frac{{qvB}}{m}{\mkern 1mu} \left( {{\mkern 1mu} \frac{{\sqrt 3 }}{2}{\mkern 1mu} \hat i + \frac{1}{2}\hat j} \right)\)
C \(\frac{{qvB}}{m}\,\left( {\,\frac{{ - \hat j + \hat i}}{{\sqrt 2 }}} \right)\)
D \(\frac{{qvB}}{m}{\mkern 1mu} \left( {\frac{1}{2}\hat j - \frac{{\sqrt 3 }}{2}\hat i} \right)\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{qvB}}{m}{\mkern 1mu} \left( {\frac{1}{2}\hat j - \frac{{\sqrt 3 }}{2}\hat i} \right)\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{B}}=\mathrm{B} \hat{\mathrm{k}}\) \(r=\frac{m v}{q B}\) \(r=2 d\) \(\sin \theta=\frac{d}{2 d} \Rightarrow \sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{6}\) \(\vec{a}=\frac{|q| v b}{m}(\cos \theta(-\hat{i})+\sin \theta \hat{j})\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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