KCET · Physics · Magnetic Effects of Current
A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is
- A \(1: 4\)
- B \(1: \sqrt{2}\)
- C \(1: 1\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(B) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
For a moving charge in a perpendicular magnetic field
\(\frac{m v^{2}}{r}=B q v \Rightarrow r=\frac{m v}{B q}=\frac{p}{B q}\)
\(\frac{r_{p}}{r_{d}}=\frac{p_{p}}{p_{d}}\)
or
Also, momentum \(p=\sqrt{2 m E}\)
(as \(q\) is same for both)
\(\frac{p_{p}}{p_{d}}=\sqrt{\frac{m_{p}}{m_{d}}}\)
From Eq. (i) and (ii), we have
\(\frac{r_{p}}{r_{d}}=\sqrt{\frac{m_{p}}{m_{d}}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)
\(\frac{m v^{2}}{r}=B q v \Rightarrow r=\frac{m v}{B q}=\frac{p}{B q}\)
\(\frac{r_{p}}{r_{d}}=\frac{p_{p}}{p_{d}}\)
or
Also, momentum \(p=\sqrt{2 m E}\)
(as \(q\) is same for both)
\(\frac{p_{p}}{p_{d}}=\sqrt{\frac{m_{p}}{m_{d}}}\)
From Eq. (i) and (ii), we have
\(\frac{r_{p}}{r_{d}}=\sqrt{\frac{m_{p}}{m_{d}}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)
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