KCET · Physics · Magnetic Effects of Current

A positively charged particle \(q\) of mass \(m\) is passed through a velocity selector. It moves horizontally righward without deviation along the line \(y=\frac{2 m v}{q B}\) with a speed \(v\). The electric field is vertically downwards and magnetic field is into the plane of paper. Now, the electric field is. switched off at \(t=0\). The angular momentum of the charged particle about origin \(O\) at \(t=\frac{\pi m}{q B}\) is

A \(\frac{m E^2}{q B^3}\)
B \(\frac{4 m^2 E^2}{q B^3}\)
C Zero
D \(\frac{m E^3}{q B^2}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{4 m^2 E^2}{q B^3}\)

Step-by-step Solution

Detailed explanation

Radius of the circle, \(r=\frac{m v}{B q}\)
\(\therefore\) Equation of line, \(y=\frac{2 m v}{B q}=2 R\)
Time pariod, \(T=\frac{2 \pi m}{B q}\)
\(t=\frac{\pi m}{B q}=\frac{T}{2}\)
\(\therefore\) Angular momentum of charged particle about origin
\(\begin{aligned} L & =m v y+m v y=2 m v y \\ & =2 m v \times 2 R \\ & =4 m v \times \frac{m v}{B q}=\frac{4 m^2 v^2}{B q}\end{aligned}\)
From velocity selector,
\(v=\frac{E}{B}\)
\(\therefore =\frac{4 m^2 E^2}{q B^3}\)

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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