/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 An electron of mass \(m\) and ch... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron of mass \(m\) and charge \(e\) is accelerated by a potential difference \(V .\) It then enters a uniform magnetic field \(B\) applied perpendicular to its path. The radius of the circular path of the electron is (A) \(r=\left(\frac{2 m V}{e B^{2}}\right)^{\frac{1}{2}}\) (B) \(r=\left(\frac{2 m e V}{B^{2}}\right)^{\frac{1}{2}}\) (C) \(r=\left(\frac{2 m B}{e V^{2}}\right)^{\frac{1}{2}}\) (D) \(r=\left(\frac{2 B^{2} V}{e m}\right)^{\frac{1}{2}}\)

Short Answer

Expert verified
The short answer is: The radius of the circular path of the electron is (A) \(r = \sqrt{\frac{2 m V}{e B^{2}}}\).

Step by step solution

01

Kinetic energy from potential difference

When the electron is accelerated through a potential difference V, it gains kinetic energy (KE). The kinetic energy can be expressed in terms of the electron charge (e) and the potential difference (V) as follows: KE = eV
02

Find the velocity of the electron

Using the relation KE = 1/2 * m * v^2, we can find the velocity (v) of the electron: eV = 1/2 * m * v^2 v^2 = (2 * e * V) / m v = \(\sqrt{\frac{2 e V}{m}}\)
03

Magnetic force on the electron

When the electron enters the magnetic field (B), perpendicular to its path, it experiences a magnetic force (F_m) given by F_m = e * v * B.
04

Centripetal force for circular motion

For the electron to maintain circular motion, the centripetal force (F_c) required is F_c = (m * v^2) / r.
05

Equate magnetic force and centripetal force

In the circular motion, the magnetic force provides the centripetal force, so we equate both forces: e * v * B = (m * v^2) / r
06

Solve for the radius, r

Substitute the expression for v from step 2: e * \(\sqrt{\frac{2 e V}{m}}\) * B = (m * \(\frac{2 e V}{m}\)) / r After simplification, we get: r = \(\sqrt{\frac{2 m V}{e B^{2}}}\) Hence, the correct option is (A) r = \(\sqrt{\frac{2 m V}{e B^{2}}}\).

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Key Concepts

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

Electron Motion
Electrons are subatomic particles with a negative charge, and their behavior in electric and magnetic fields is foundational in electromagnetism. When electrons move, especially due to an external influence like a potential difference, their velocity changes, leading to new dynamics. In physics, that movement is highly influenced by factors such as electric fields, described by potential differences, and magnetic fields. In this scenario, an electron starts by being accelerated from rest by a potential difference (voltage). This acceleration results in the electron gaining speed as it converts potential energy into kinetic energy. The way electrons gain speed and energy is similar to a car accelerating until it reaches a cruising speed on a highway. As they speed up, their motion can be influenced further by other forces and fields, such as magnetic fields, causing changes in their path.
Magnetic Fields
Magnetic fields are regions in space where magnetic forces can be detected. These fields arise from magnets or the flow of electric currents. For an electron, interacting with a magnetic field is an everyday event in devices like TV screens or MRI machines.When an electron with a velocity enters a magnetic field, it experiences a force called the magnetic force. This force is always perpendicular to both the velocity of the electron and the magnetic field itself, changing the path of the electron.
  • The magnetic force is given by: \[ F_m = e \cdot v \cdot B \]
  • - Here, \( e \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field strength.
When applied perpendicular to the motion of the electron, this force can cause the electron to move in a curved path, often circular, thus turning the straight motion into a circular one.
Centripetal Force
To keep an electron in a circular path, a force must act continuously towards the center of the circle. This force is known as the centripetal force. It is the same kind of force that keeps the Earth orbiting the Sun or a roller coaster car traveling around a loop without falling.In the context of electron motion in a magnetic field, the magnetic force acts as the centripetal force. Therefore, these two forces are equal in magnitude but point in different directions:- The mathematical representation of centripetal force is: \[ F_c = \frac{m \cdot v^2}{r} \]- Here, \( m \) is the mass of the electron, \( v \) is its velocity, and \( r \) is the radius of the circular path.So, the magnetic force is essentially providing the necessary centripetal force to keep the electron on its curved path. By setting them equal, physicists can determine the characteristics of the electron's path, like its radius.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For electrons, kinetic energy determines how fast they move, which affects how they interact with electromagnetic fields.In this exercise, an electron gains kinetic energy by being accelerated through a potential difference. The relationship between electron charge \( e \) and potential difference \( V \) gives us an expression for the kinetic energy:- The formula for this is: \[ KE = e \cdot V \]This equation provides the stepping stone to find other properties like velocity. From the kinetic energy, using the relationship:\[ KE = \frac{1}{2}m v^2 \]we can rearrange the terms to solve for the velocity of the electron: \[ v = \sqrt{\frac{2 e V}{m}} \]Understanding kinetic energy is crucial because it connects potential differences (what pushes electrons) with their speed and movement, ultimately affecting their path in magnetic fields.

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Most popular questions from this chapter

In hydrogen like atom an electron is orbiting in an orbit having quantum number \(n\). Its frequency of revolution is found to be \(13.2 \times 10^{15} \mathrm{~Hz}\). Energy required to free the electron from the atom from the above orbit is \(54.4 \mathrm{eV}\). In time 7 nano second the electron jumps back to orbit having quantum number \(\frac{n}{2} . \tau\) be the average torque acted on the electron during the above process, then find \(\tau \times 10^{27}\) in \(\mathrm{Nm}\). (given : \(\frac{h}{\pi}=\) \(2.1 \times 10^{-34} \mathrm{~J}-\mathrm{s}\), frequency of revolution of electron in the ground state of H-atom \(v_{0}=6.6 \times 10^{15} \mathrm{~Hz}\) and ionization energy of \(\mathrm{H}\)-atom, \(E_{0}=13.6 \mathrm{eV}\) ).

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its \(K E\). Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum \(K E\) of colliding electron is (A) \(10.2 \mathrm{eV}\) (B) \(1.9 \mathrm{eV}\) (C) \(12.09 \mathrm{eV}\) (D) \(13.6 \mathrm{eV}\)

When \({ }_{3} \mathrm{Li}^{7}\) nuclei are bombarded by protons, and the resultant nuclei are \({ }_{4} \mathrm{Be}^{8}\), the emitted particles will be (A) Neutrons. (B) Alpha particles. (C) Beta particles. (D) Gamma photons.

The ratio of de Broglie wavelength of \(\alpha\)-particle to that of a proton being subjected to the same magnetic field so that the radii of their paths are equal to each other assuming the field induction vector \(\vec{B}\) is perpendicular to the velocity vectors of the \(\alpha\)-particle and the proton is (A) 1 (B) \(\frac{1}{4}\) (C) \(\frac{1}{2}\) (D) 2

Two radioactive substances \(X\) and \(Y\) initially contain equal number of nuclei. \(X\) has a half-life of 1 hour and \(Y\) has half-life of 2 hours. After two hours the ratio of the activity of \(X\) to the activity of \(Y\) will be (A) \(1: 4\) (B) \(1: 2\) (C) \(1: 1\) (D) \(2: 1\)

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