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Chapter 13: Problem 1908

A proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.

Short Answer

Expert verified
The radius of the path of the alpha particle will be greater than that of the proton (option b).

Step by step solution

01

Write down the given information

We are given that both the proton and the alpha particle have the same kinetic energy and are projected at right angles to a uniform magnetic field. Let's denote the kinetic energy as \(K\) and the magnitude of the magnetic field as \(B\).
02

Express the velocity of each particle in terms of kinetic energy

We can use the formula for kinetic energy to write the velocity of each particle in terms of the kinetic energy. The kinetic energy formula is given by \(K = \frac{1}{2} m v^2\), where \(m\) is the mass of the particle and \(v\) its velocity. Solving for \(v\), we have: \[v = \sqrt{\frac{2K}{m}}\]
03

Calculate the radius for each particle

For the proton, we have mass \(m_p = 1.67 \times 10^{-27}\, kg\) and charge \(q_p = 1.6 \times 10^{-19}\, C\). For the alpha particle, we have mass \(m_{\alpha} = 6.64 \times 10^{-27}\, kg\) and charge \(q_{\alpha} = 3.20 \times 10^{-19}\, C\). Using the formula for the radius in a magnetic field, we have: For the proton: \[r_p = \frac{m_p v_p}{q_p B} = \frac{m_p \sqrt{\frac{2K}{m_p}}}{q_p B}\] For the alpha particle: \[r_{\alpha} = \frac{m_{\alpha} v_{\alpha}}{q_{\alpha} B} = \frac{m_{\alpha} \sqrt{\frac{2K}{m_{\alpha}}}}{q_{\alpha} B}\]
04

Compare the radius of the two particles

Using the expressions for \(r_p\) and \(r_{\alpha}\), we can compare their radius: \[\frac{r_{\alpha}}{r_p} = \frac{\frac{m_{\alpha} \sqrt{\frac{2K}{m_{\alpha}}}}{q_{\alpha} B}}{\frac{m_p \sqrt{\frac{2K}{m_p}}}{q_p B}} = \frac{m_{\alpha} q_p}{m_p q_{\alpha}}\] Since the alpha particle has twice the charge and four times the mass of the proton, this fraction becomes: \[\frac{r_{\alpha}}{r_p} = \frac{4m_p \times q_p}{m_p \times 2q_p} = 2\] This tells us that the radius of the path of the alpha particle is greater than that of the proton.
05

Choose the correct answer

We have found that the radius of the path of the alpha particle is greater than that of the proton. This corresponds to option (b), which is the correct answer.

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Most popular questions from this chapter

A He nucleus makes a full rotation in a circle of radius \(0.8\) meter in $2 \mathrm{sec}\(. The value of the mag. field \)\mathrm{B}$ at the centre of the circle will be \(\quad\) Tesla. (a) \(\left(10^{-19} / \mu_{0}\right)\) (b) \(10^{-19} \mu_{0}\) (c) \(2 \times 10^{-10} \mathrm{H}_{0}\) (d) \(\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]\)

Two concentric co-planar circular Loops of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) carry currents of respectively \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) in opposite directions. The magnetic induction at the centre of the Loops is half that due to \(\mathrm{I}_{1}\) alone at the centre. If \(\mathrm{r}_{2}=2 \mathrm{r}_{1}\) the value of $\left(\mathrm{I}_{2} / \mathrm{I}_{1}\right)$ is (a) 2 (b) \(1 / 2\) (c) \(1 / 4\) (d) 1

A conducting rod of 1 meter length and \(1 \mathrm{~kg}\) mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$ (a) \(0.5 \mathrm{Amp}\) (b) \(15 \mathrm{Amp}\) (c) \(5 \mathrm{Amp}\) (d) \(1.5 \mathrm{Amp}\)

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of $2.5 \mathrm{Amp}$. The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

A straight wire of length \(30 \mathrm{~cm}\) and mass 60 milligram lies in a direction \(30^{\circ}\) east of north. The earth's magnetic field at this site is horizontal and has a magnitude of \(0.8 \mathrm{G}\). What current must be passed through the wire so that it may float in air ? \(\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (a) \(10 \mathrm{Amp}\) (b) \(20 \mathrm{Amp}\) (c) \(40 \mathrm{Amp}\) (d) \(50 \mathrm{Amp}\)
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