Question

In an experiment with cosmic rays, a vertical beam of particles that have charge of magnitude 3e and mass 12 times the proton mass enters a
uniform horizontal magnetic field of 0.250 T and is bent in a semicircle of diameter 95.0 cm, as shown in the figure.
a) Find the speed of the particles.
b) Find the sign of particles' charge.
c) Is it reasonable to ignore the gravity force on the particles?
d) How does the speed of the particles as they enter the field compare to their speed as they exit the field?

Answer 1

Concepts and reason

The concept of the magnetic force exerted on a moving charged particle through a magnetic field and the centripetal force on a particle moving in a circle are required to solve the problem. First, calculate the speed of the particles by equating the expression of Lorentz force and centripetal force. The sign of the charged particles is determined by using the right-hand thumb rule. Then, determine the magnetic force by using the Lorentz force law and compare the force of gravity and magnetic force by using the magnitude of the force of gravity and magnetic force. Finally, compare the speed of the particles as they enter the field and as they exit the field by finding the work done by the magnetic field.

Fundamentals

Centripetal force is the force that causes an object to move along a circular path. The expression for centripetal force is, \(F=\frac{m v^{2}}{r}\)

Here, \(m\) is the mass of the object, \(v\) is the velocity of the object, and \(r\) is the radius of the circular path. The magnetic force exerted on a charged particle moving with velocity \(v\) through a magnetic field \(B\) is given by the following relation:

\(F=q v B \sin \theta\)

Here, \(\mathrm{q}\) is the charge of the particle and \(\theta\) is the angle between the velocity and magnetic field. The right-hand thumb rule is used to find the direction of the magnetic force. Initially, show the fingers of the right hand in the direction of velocity. Later curl them towards the direction of the magnetic field. Then, the thumb points in the direction of the magnetic force. The force on an object due to gravity \(F_{\mathrm{W}}\) acts in a downward direction and it is determined by using the following relation:

\(F_{\mathrm{W}}=m g\)

Here, \(\mathrm{m}\) is the mass of the object and \(\mathrm{g}\) is the acceleration due to gravity.

(a) Substitute \(\frac{m v^{2}}{r}\) for \(\mathrm{F}\) and \(90^{\circ}\) for \(\theta\) in equation \(F=q v B \sin \theta\) and solve for \(\mathrm{v}\).

$$ \begin{array}{c} \frac{m v^{2}}{r}=q v B \sin 90^{\circ} \\ v=\frac{r q B}{m} \end{array} $$

Substitute \(\frac{d}{2}\) for \(r, 3 e\) for \(q\), and \(12 m_{\mathrm{p}}\) for \(m\) in equation \(v=\frac{r q B}{m}\). \(v=\frac{d e B}{8 m_{\mathrm{p}}}\)

Here, e is the unit electric charge, \(\mathrm{d}\) is the diameter of the circular path, and \(m_{\mathrm{p}}\) is the mass of the proton. Substitute \(95.0 \mathrm{~cm}\) for d, \(1.6 \times 10^{-19} \mathrm{C}\) for e, \(1.67 \times 10^{-27} \mathrm{~kg}\) form \(_{\mathrm{p}}\), and \(0.25 \mathrm{~T}\) for \(\mathrm{B}\) in the equation \(v=\frac{d e B}{8 m_{\mathrm{p}}}\).

$$ \begin{array}{c} v=\frac{\left(95 \mathrm{~cm}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)(0.25 \mathrm{~T})}{8\left(1.67 \times 10^{-27} \mathrm{~kg}\right)} \\ =2.84 \times 10^{6} \mathrm{~m} / \mathrm{s} \end{array} $$

Hence, the speed of the particles is \(2.84 \times 10^{6} \mathrm{~m} / \mathrm{s}\).

The direction of the magnetic force acting on the moving particle is always perpendicular to the motion of the particle. Hence, the magnetic force tries to move the particle in a circular path. The magnetic field is perpendicular to the velocity of the particles. Therefore, the value of \(\theta\) is equal to \(90^{\circ}\).

(b) Determine the sign of the particle's charge. By using the right-hand rule, the thumb shows the opposite direction to the centripetal force. This implies that the particle is negatively charged.

The sign of the charge can be identified by using the right-hand rule. If the particle is negatively charged, then the centripetal force acts in the direction opposite to the direction of thumb points. On the other hand, if the particles are positively charged, then the centripetal force acts in the direction of thumb points.

(c) Calculate the gravitational force acting on the particle. The mass of the particles is, \(m=12 m_{\mathrm{p}}\)

Here, \(m_{\mathrm{p}}\) is the mass of the proton. Substitute \(1.67 \times 10^{-27} \mathrm{~kg}\) for \(m_{\mathrm{p}}\) in equation \(m=12 m_{\mathrm{p}}\) and calculate the mass of the particle.

$$ \begin{aligned} m &=12\left(1.67 \times 10^{-27} \mathrm{~kg}\right) \\ &=2.004 \times 10^{-26} \mathrm{~kg} \end{aligned} $$

The force of gravity acting on the particles is, \(F_{\mathrm{W}}=m g\)

Here, \(\mathrm{m}\) is the mass of the particles. Substitute \(2.004 \times 10^{-26} \mathrm{~kg}\) for \(\mathrm{m}\) and \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}\) in equation \(F_{\mathrm{W}}=m g\)

The gravitational force acting on the particle is the force due to gravity and it is equal to the product of the mass of the particle and acceleration due to gravity. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{~kg}\). The value of acceleration due to gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)

Compare the gravitational force and magnetic force acting on the particle. The charge on the particle is, \(q=3 e\)

Here, e is the unit electric charge. Substitute \(1.6 \times 10^{-19} \mathrm{C}\) for e in equation \(q=3 e\) and determine the magnitude of the charge on the particle.

$$ \begin{aligned} q=& 3\left(1.6 \times 10^{-19} \mathrm{C}\right) \\ &=4.8 \times 10^{-19} \mathrm{C} \end{aligned} $$

Substitute \(90^{\circ}\) for \(\theta, 4.8 \times 10^{-19} \mathrm{C}\) for \(\mathrm{q}, 2.84 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(\mathrm{v}\), and \(0.25 \mathrm{~T}\) for \(\mathrm{B}\) in equation \(F=q v B \sin \theta\)

$$ F=\left(4.8 \times 10^{-19} \mathrm{C}\right)\left(2.84 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)(0.25 \mathrm{~T}) \sin 90^{\circ} $$

$$ =3.41 \times 10^{-13} \mathrm{~N} $$

Compare the magnetic force (F) and gravitational force \(\left(F_{\mathrm{W}}\right)\).

$$ \frac{F}{F_{\mathrm{W}}}=\frac{3.41 \times 10^{-13} \mathrm{~N}}{1.96 \times 10^{-25} \mathrm{~N}} $$

\(F=\left(1.74 \times 10^{12}\right) F_{\mathrm{W}}\)

From this equation, the magnetic force is in order of \(10^{12}\) times the force due to gravity. Hence, the force due to gravity can be neglected.

The magnetic field is perpendicular to the velocity of the particles. Therefore, the angle between the velocity and magnetic field is \(90^{\circ}\). The value of the unit electric charge is \(1.6 \times 10^{-19} \mathrm{C}\). The magnetic force is \(1.74 \times 10^{12}\) times greater than the force due to gravity which implies that the magnetic force is very large as compared with the force due to gravity. Hence, the force due to gravity can be neglected.

(d) The force acting on the particle is perpendicular to the velocity of the particle which implies no displacement in the direction of the force. Hence, the work done on the particle is zero. As the work is done on the particle is zero, the speed of the particle remains the same. Hence, the speed of the particles when they enter the field is the same as the speed when they exit the field.

Work is done by the force if the force acting on the particle causes it to displace in the direction of the force. If there is no displacement in the direction of force then the work done is zero.

Part a The speed of the particles is \(2.84 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Part b The sign on the charged particles is negative.

Part c It is reasonable to neglect the force due to gravity.

Part d The speed of the particles as they enter the field is same as the speed as they exit the field.