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A charged particle of mass \(m\) and charge \(q\) is projected into a uniform magnetic field of induction \(B\) with speed \(v\) which is perpendicular to \(B\). The width of the magnetic field is \(d\). the impulse imparted to the particle by the field is \((d<

Short Answer

Expert verified
Given the conditions of the problem, none of the provided choices (A - D) fit the final calculated value of impulse to a charged particle moving in the magnetic field. The closest approximation would be (C) \( qBd \), but this is not exactly accurate to the calculation.

Step by step solution

01

Calculate the Radius

When a charged particle is projected into a magnetic field perpendicular to the field, it follows a circular path. The radius (r) of this path is given by the equation \( r = \frac{mv}{qB} \). This is calculated from the balance between the centripetal force \( \frac{mv^2}{r} \) and the magnetic Lorentz force \( qvB \) on the particle in the magnetic field.
02

Calculate the Circumference

Next, calculate the circumference of the circular path. This is the total distance that the particle travels in the magnetic field. The formula for the circumference of a circle is \( C = 2Ï€r \). Using the value of r from Step 1, \( C = 2Ï€ \frac{mv}{qB} \). Remembering that the width of the magnetic field is d, and given that \( d<< C \), the total path covered by the particle through the magnetic field is approximated as \( d = C \).
03

Calculate the Impulse

The definition of impulse is the change in linear momentum. In our case, because the direction of velocity changes but magnitude remains same due to circular path, the change in momentum is the difference between the final and initial momentum vectors, and is double of the any one. Therefore, the linear momentum of the particle by moving over a path of length \( d = C \) in the field can be calculated as \( Impulse = 2 \times mv = 2 \times m \times \frac{qBd}{2Ï€} = \frac{mqBd}{Ï€} \).
04

Approximate the value of π

Approximate the value of π as 3 (for simplicity in physics computations) and hence the Impulse can be simplified as \( \frac{mqBd}{3} \).

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Key Concepts

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

Centripetal Force
When an object moves in a circular path, whether it's a planet orbiting a star or a charged particle in a magnetic field, it constantly changes direction. This change in direction requires a force. The force that keeps an object moving in a circular path is called the centripetal force. The centripetal force is always directed towards the center of the circle. It does not change the speed of the object but constantly changes its direction to keep it in a circular motion. The formula for centripetal force is:\[ F_c = \frac{mv^2}{r} \]Here, \( m \) denotes the mass of the object, \( v \) represents its velocity, and \( r \) is the radius of the circular path. This formula helps us understand how the mass, speed, and radius affect the required centripetal force.- An increase in speed or a decrease in radius demands a greater centripetal force to maintain circular motion. - It is essential for stability in systems involving rotational dynamics, including planetary orbits and even vehicles taking turns on roads.
Lorentz Force
The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. In our context, we're focusing on the magnetic part of it. This force is crucial because it can change the direction of the particle without affecting its speed, leading to circular motion. The magnetic component of the Lorentz force is given by:\[ F_m = qvB \]Where:- \( q \) is the charge of the particle - \( v \) is the speed of the particle - \( B \) is the magnetic field strengthThis force acts perpendicular to both the velocity of the particle and the direction of the magnetic field. As a result, it causes the particle to move in a circular path. This force is what provides the necessary centripetal force to keep the particle on this path. - Remember, the stronger the magnetic field, the greater the Lorentz force. - The magnitude of the charge and velocity also play a significant role in determining the force experienced.
Circular Motion
Circular motion occurs when an object travels along a circular path at a constant speed. This is a common occurrence in both natural phenomena and engineered systems. In the case of a charged particle in a magnetic field, like our exercise, the interaction of the Lorentz force and the particle's velocity creates this motion. The radius of the circular motion is determined by balancing the magnetic Lorentz force with the required centripetal force:\[ r = \frac{mv}{qB} \]- A larger radius is associated with a larger mass or smaller charge. - A stronger magnetic field or faster speed can lead to a tighter circular path.Though the speed remains constant in circular motion, the direction is always changing, which results in the changing velocity vector. This is why acceleration (centripetal acceleration) is always present, even though speed does not change. Understanding circular motion is essential because it underpins many physical systems, from the dynamics of galaxies to the design of particle accelerators.

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Most popular questions from this chapter

In a uniform magnetic field of induction \(B\), a wire in the form of semicircle of radius \(r\) rotates about the diameter of the circle with angular frequency \(\omega\). If the total resistance of the circuit is \(R\), the mean power generated per period of rotation is (A) \(\frac{B \pi r^{2} \omega}{2 R}\) (B) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{2 R}\) (C) \(\frac{(B \pi r \omega)^{2}}{2 R}\) (D) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\)

A positive charge is passing through an electromagnetic field in which \(\vec{E}\) and \(\vec{B}\) are directed towards \(y\)-axis and \(z\)-axis, respectively. If a charged particle passes through the region undeviated, then its velocity is/are represented by (here \(a, b\), and \(c\) are constant) (A) \(\vec{v}=\frac{E}{B} \hat{i}+a \hat{j}\) (B) \(\vec{v}=\frac{E}{B} \hat{i}+b \hat{k}\) (C) \(\vec{v}=\frac{E}{B} \hat{i}+c \hat{i}\) (D) \(\vec{v}=\frac{E}{B} \hat{i}\)

A bar magnet, of magnetic moment \(M\), is placed in a magnetic field of induction \(B\). The torque exerted on it is (A) \(\vec{M} \cdot \vec{B}\) (B) \(\vec{B} \times \vec{M}\) (C) \(\vec{M} \times \vec{B}\) (D) \(-\vec{B} \cdot \vec{M}\)

Two particles each of mass \(m\) and charge \(q\) are attached to the two ends of a light rigid rod of length \(2 \ell\). The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) \(\frac{q}{2 m} \quad\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{q}{\pi m}\)

If a coil of metal wire is kept stationary in a non-uniform magnetic field, (A) An EMF and current are both induced in the coil (B) A current but no EMF is induced in the coil (C) An EMF but no current is induced in the coil (D) Neither EMF nor current is induced in the coil

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