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Chapter 15: Problem 150

If an electron and a proton having same momenta enter perpendicular to a magnetic field, then (A) curved path of electron and proton will be same (ignoring the sense of revolution). [2002] (B) they will move undeflected. (C) curved path of electron is more curved than that of the proton. (D) path of proton is more curved.

Short Answer

Expert verified
(C) The curved path of the electron is more curved than that of the proton.

Step by step solution

01

Determine the magnetic force on moving charges

Using the formula for the magnetic force on a moving charge, we can write \[F = qvB \sin\theta\] where F is the magnetic force, q is the charge, v is the velocity of the charged particle, B is the magnetic field, and \(\theta\) is the angle between the velocity and the magnetic field direction. In our case, \(\theta = 90^\circ\), so \(\sin\theta = 1\). Thus, the magnetic force becomes \[F = qvB\]
02

Apply the centripetal force formula

The magnetic force acting on the particle causes the particle to move in a circular path. The centripetal force needed to keep the particle in a circular path can be written as \[F_c = \frac{mv^2}{R}\] where \(F_c\) is the centripetal force, m is the mass of the particle, v is the velocity of the particle, and R is the radius of curvature of the particle's path.
03

Equate the magnetic force and centripetal force

Since the magnetic force is responsible for keeping the charged particle in a circular path, we can equate the magnetic force and the centripetal force as follows: \[qvB = \frac{mv^2}{R}\]
04

Determine the radius of curvature

Now, we will solve the equation from step 3 to find the radius of curvature R for each particle: \[R = \frac{mv}{qB}\]
05

Apply the same momentum for both particles

Since the electron and proton have the same momentum, we can write \[m_ev_e = m_pv_p\] where \(m_e\) and \(m_p\) are the masses of the electron and proton, respectively, and \(v_e\) and \(v_p\) are their velocities, respectively.
06

Find the ratio of radii

Now, we will find the ratio of radii of the curved paths of the electron and proton using the equations from steps 4 and 5: \[\frac{R_e}{R_p} = \frac{m_ev_e}{q_eB} \times \frac{q_pB}{m_pv_p} = \frac{m_ev_e}{q_e} \times \frac{q_p}{m_pv_p}\] Using the same momentum condition from step 5, substitute \(m_ev_e = m_pv_p\) and simplify: \[\frac{R_e}{R_p} = \frac{m_e}{q_e} \times \frac{q_p}{m_p}\] Since the charge of the electron and proton are equal, \(|q_e| = |q_p|\) and we have: \[\frac{R_e}{R_p} = \frac{m_e}{m_p}\]
07

Compare the radii

Since the mass of an electron (\(m_e\)) is much smaller than the mass of a proton (\(m_p\)), we can conclude that the radius of curvature of the electron's path (\(R_e\)) is also much smaller than the radius of curvature of the proton's path (\(R_p\)). Therefore, the correct answer is: (C) The curved path of the electron is more curved than that of the proton.

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