આકૃતિમાં દર્શાવ્યા પ્રમાણે એક સમક્ષિતિજ સાથે \(\theta\) ખૂણે અને \(l\) અંતરે રહેલ પાટા પર \(m\) દળનો સળિયો ગુરુત્વાકર્ષણની અસર હેઠળ ગતિ કરે છે. પાટાનાં તળિયે એક \(R\) અવરોધ જોડેલો છે. પાટાનાં સમતલને લંબ એકસમાન ચુંબકીયક્ષેત્ર \(B\) છે. તો કોપરના સળિયાનો ટર્મિનલ વેગ કેટલો થાય?

A \(\frac{{mgR\cos \,\theta }}{{{B^2}{l^2}}}\)
B \(\frac{{mgR\sin \,\theta }}{{{B^2}{l^2}}}\)
C \(\frac{{mgR\tan \,\theta }}{{{B^2}{l^2}}}\)
D \(\frac{{mgR\cot \,\theta }}{{{B^2}{l^2}}}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{{mgR\sin \,\theta }}{{{B^2}{l^2}}}\)

Step-by-step Solution

Detailed explanation

From Faraday's law of electomagnetic induction, \(e=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t}\) \( = \frac{{Bdl \times l}}{{dt}} = BVl\) Also, \(F=i l B=\left(\frac{B V}{R}\right)\left(l^{2} B\right)=\frac{B^{2} l^{2} V}{R}\) At equilibrium…

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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