WBJEE · Physics · Electrostatics
A light charged particle is revolving in a circle of radius \(r\) in electrostatic attraction of a static heavy particle with opposite charge How does the magnetic field vary at the centre of the circle due to the moving charge depend on \(r ?\)
- A \(B \propto \frac{1}{r}\)
- B \(B \propto \frac{1}{r^{2}}\)
- C \(B \propto \frac{1}{r^{\frac{3}{2}}}\)
- D \(B \propto \frac{1}{r^{\frac{5}{2}}}\)
Answer & Solution
Correct Answer
(D) \(B \propto \frac{1}{r^{\frac{5}{2}}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} &\begin{array}{l} F_{\text {centripetal }}=m r \omega^{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}} \\ \therefore \quad \omega^{2}=\frac{1}{4 \pi \varepsilon_{0} m} \cdot \frac{q_{1} q_{2}}{r^{3}} \Rightarrow \omega=\sqrt{\frac{q_{1}…
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