AP EAMCET · PHYSICS · Magnetic Effects of Current
A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loses half of its kinetic energy, then the radius of curvature of its path is
- A No change
- B Reduced by \(\frac{1}{2}\) times of its initial values
- C Reduced to \(\frac{1}{\sqrt{2}}\) times of its initial values
- D Reduce to \(\frac{1}{4}\) times of its initial values
Answer & Solution
Correct Answer
(C) Reduced to \(\frac{1}{\sqrt{2}}\) times of its initial values
Step-by-step Solution
Detailed explanation
We know that \(r=\frac{m V}{q B}=\frac{\sqrt{2 m(K . E)}}{q B}\)…
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