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Chapter 13: Problem 1906

Two particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

Short Answer

Expert verified
The ratio of the mass of particle X to that of particle Y is given by: \[\frac{m_x}{m_y} = \frac{r_2}{r_1}\] which corresponds to the option (b).

Step by step solution

01

Determining the velocity of the particles using the potential difference

The kinetic energy gained by the particles after being accelerated through the potential difference V is given by: \[K.E. = qV\] where q is the charge of the particles. The kinetic energy of a particle is also given by: \[K.E. = \frac{1}{2}m\upsilon^2\] where m is the mass of the particle and 饾湀 is its velocity. Equating both expressions for kinetic energy, we can determine the velocity of the particles as follows: \[qV = \frac{1}{2}m\upsilon^2\]
02

Apply the Lorentz force formula to the particles

The Lorentz force acting on a charged particle in a magnetic field is given by: \[F = q\upsilon B\] where B is the magnetic field. Since the charged particles are moving in a circular path, the centripetal force required is given by: \[F_c = \frac{m\upsilon^2}{r}\] Equating the Lorentz force with the centripetal force, we get: \[q\upsilon B = \frac{m\upsilon^2}{r}\]
03

Finding the ratio of the masses

We need to find the ratio of the masses of the particles X and Y. Let's denote these masses as 饾憵饾懃 and 饾憵饾懄, respectively. We have the following equations for the particles X and Y from step 2: \[q\upsilon_x B = \frac{m_x\upsilon_x^2}{r_1}\] \[q\upsilon_y B = \frac{m_y\upsilon_y^2}{r_2}\] Now, 饾憺 and 饾惖 are equal for both particles, so we can divide the first equation by the second equation to get the ratio of the masses: \[\frac{m_x\upsilon_x^2/r_1}{m_y\upsilon_y^2/r_2} = 1\] Rearranging the equation to find the ratio (m_x/m_y), we get: \[\frac{m_x}{m_y} = \frac{\upsilon_x^2/r_1}{\upsilon_y^2/r_2}\] Now, using the formula for the velocity we derived in step 1, substitute the velocities in terms of potential difference: \[\upsilon_x^2 = \frac{2qV}{m_x}\] \[\upsilon_y^2 = \frac{2qV}{m_y}\] Substitute these values into the ratio equation: \[\frac{m_x}{m_y} = \frac{\left(\frac{2qV}{m_x}\right)/r_1}{\left(\frac{2qV}{m_y}\right)/r_2}\] Simplify the equation: \[\frac{m_x}{m_y} = \frac{2qVr_2}{2qVr_1}\] The term \(2qV\) cancels out from both numerator and denominator, leaving us with the final answer: \[\frac{m_x}{m_y} = \frac{r_2}{r_1}\] So, the ratio of the mass of particle X to that of particle Y is given by: \[\frac{m_x}{m_y} = \frac{r_2}{r_1}\] which corresponds to the option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

charged particles
Charged particles, like electrons and protons, possess either a positive or negative electric charge. This charge is a fundamental property that dictates how they interact with electric and magnetic fields.
  • When a charged particle moves through a magnetic field, it experiences a force called the Lorentz force, which causes it to change direction.
  • This reaction is crucial in many scientific and technological applications, such as particle accelerators and even in medical imaging technologies like MRI.
  • In the context of this exercise, particles X and Y have equal charges, which means they will experience similar initial forces when subjected to the magnetic field.
Understanding how charged particles behave in different fields lays the groundwork for exploring their motion when placed within a uniform magnetic field, as we'll examine next.
circular motion
When a charged particle enters a magnetic field at an angle, it tends to move in a circular path. This circular motion is a result of the continuous change in the direction of the velocity of the particle but not in its speed.
  • The constant Lorentz force exerted by the magnetic field keeps the particle moving in a circular trajectory.
  • The radius of this circular motion depends on several factors, including the mass of the particle, its charge, its velocity, and the strength of the magnetic field.
  • In our exercise, particles X and Y describe circular paths with radii \( r_1 \) and \( r_2 \). These path radii are essential for determining the mass ratio of the two particles.
Understanding circular motion helps explain how magnetic fields control charged particles, which is central to many technologies, from cyclotrons for nuclear physics research to devices in everyday electronics.
centripetal force
Centripetal force is a fundamental concept when discussing motion in a circle. It is the force that is required to keep an object moving in a circular path, acting towards the center of the curvature of the path.
  • In case of charged particles moving in a magnetic field, the centripetal force is provided by the Lorentz force.
  • This force keeps the charged particle from moving off in a straight line, redirecting it into circular motion.
  • The centripetal force needed is given by \( F_c = \frac{m\upsilon^2}{r} \), where \( m \) is the mass of the particle, \( \upsilon \) its velocity, and \( r \) the radius of the circle.
Balancing this force with the magnetic force, \( F = q\upsilon B \), allows the calculation of parameters like mass ratios, velocity, and path radius of particles in fields, as demonstrated in our solution.
uniform magnetic field
A uniform magnetic field is one in which the magnetic field strength and direction are constant throughout a given space. This consistency allows for predictable and uniform behavior of charged particles entering the field.
  • When charged particles move through a uniform magnetic field, they experience a constant force perpendicular to their velocity.
  • This force leads to a consistent speed and a change in direction, resulting in circular or helical motion, depending on the initial angle of entry into the field.
  • The properties of the field allow for calculations like those in our exercise, where the radii of the paths taken by the particles can reveal details about their mass.
Overall, uniform magnetic fields are vital in research and industry because they provide the stable conditions needed to analyze and utilize particle behavior.

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Most popular questions from this chapter

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

Needles \(\mathrm{N}_{1}, \mathrm{~N}_{2}\) and \(\mathrm{N}_{3}\) are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will (a) Attract \(\mathrm{N}_{1}\) strongly, \(\mathrm{N}_{2}\) weakly and repel \(\mathrm{N}_{3}\) weakly (b) Attract \(\mathrm{N}_{1}\) strongly, but repel \(\mathrm{N}_{2}\) and \(\mathrm{N}_{3}\) weakly (c) Attract all three of them (d) Attract \(\mathrm{N}_{1}\) and \(\mathrm{N}_{2}\) strongly but repel \(\mathrm{N}_{2}\)

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

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

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other. 3 Amp and \(4 \mathrm{Amp}\) are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be Tesla. (a) \(5 \times 10^{-5}\) (b) \(7 \times 10^{-5}\) (c) \(12 \times 10^{-5}\) (d) \(10^{-5}\)
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