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Chapter 15: Problem 15

A charged particle moves in a uniform magnetic field of induction \(\vec{B}\) with a velocity \(\vec{v}\). The change in kinetic energy in the magnetic field is zero when the velocity \(\vec{v}\) is (A) parallel to \(\vec{B}\) (B) perpendicular to \(\vec{B}\) (C) at any angle to \(\vec{B}\) (D) None of these

Short Answer

Expert verified
The change in kinetic energy in the magnetic field is zero when the velocity vector \(\vec{v}\) is parallel to \(\vec{B}\) (Option A).

Step by step solution

01

Magnetic Force Formula

The magnetic force, F, acting on a charged particle moving with velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\) can be calculated using the equation: \[ \vec{F} = q(\vec{v} \times \vec{B})\] Here, \(q\) is the charge of the particle, \(\times\) represents the cross product between velocity and magnetic field vectors.
02

Work-Energy Principle

According to the work-energy principle, the change in kinetic energy of a particle equals the work done on the particle by external forces. Mathematically, \[ W = \Delta KE\] Here, \(W\) is the work done by the magnetic force and \(\Delta KE\) is the change in the particle's kinetic energy.
03

Relating Magnetic Force to Work Done

The work W done by a force \(\vec{F}\) on an object as it moves a displacement \(\vec{d}\) can be calculated using the dot product between force and displacement, i.e. \[ W = \vec{F} \cdot \vec{d}\]
04

Zero Change in Kinetic Energy

Since we want the change in kinetic energy in the magnetic field to be zero, we want the work done by the magnetic force to be zero, i.e., \(W = 0\). As a result: \[ 0 = \vec{F} \cdot \vec{d}\] Recall that the magnetic force, \(\vec{F}\), is given by \[ \vec{F} = q(\vec{v} \times \vec{B})\] So, substituting this relation, we have: \[ 0 = q(\vec{v} \times \vec{B}) \cdot \vec{d}\]
05

Condition for Zero Work Done

Now, we have the relation for zero work done: \[ 0 = q(\vec{v} \times \vec{B}) \cdot \vec{d}\] Recall that for any two vectors A and B, the dot product of A cross B with another vector C is given by: \[(\vec{A} \times \vec{B}) \cdot \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}\] Using this identity for the vectors \(\vec{v}\), \(\vec{B}\), and \(\vec{d}\), we find: \[ 0 = (q(\vec{v} \cdot \vec{d})\vec{B} - q(\vec{B} \cdot \vec{d})\vec{v})\] The only way for this expression to be equal to zero in all cases is if the two terms cancel each other out, which means that either \[ \vec{v} \cdot \vec{d} = 0\] or \[ \vec{B} \cdot \vec{d} = 0\] In either case, the cross product term, i.e., \(\vec{v} \times \vec{B} = 0\). Since the cross product is zero when the two vectors are parallel (i.e., they point in the same or opposite direction) or one of the vectors is zero: Therefore, the change in kinetic energy in the magnetic field is zero when the velocity vector \(\vec{v}\) is (A) parallel to \(\vec{B}\)

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Key Concepts

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

Kinetic Energy
Kinetic energy refers to the energy that a particle or object possesses due to its motion. The formula to calculate kinetic energy (KE) is given by:\[KE = \frac{1}{2}mv^2\]where:
  • \(m\): Mass of the object
  • \(v\): Velocity of the object
In the context of a charged particle in a magnetic field, changes in kinetic energy are crucial to understand. However, the fundamental aspect of kinetic energy in this situation is that magnetic forces do not perform work. As a result, they do not alter a particle's kinetic energy. Therefore, a charged particle's speed remains unchanged when moving through a magnetic field as long as no other forces are acting on it.
Cross Product
The cross product is a mathematical operation between two vectors, resulting in a third vector perpendicular to the plane defined by the original vectors. The magnitude of this vector is given by:\[\|\vec{A} \times \vec{B}\| = \|\vec{A}\|\|\vec{B}\|\sin(\theta)\]where:
  • \(\|\vec{A}\|, \|\vec{B}\|\): Magnitudes of the vectors
  • \(\theta\): Angle between the vectors
In magnetic force calculations, the cross product between the velocity vector \(\vec{v}\)and the magnetic field \(\vec{B}\)gives us a vector representing the magnetic force \(\vec{F}\). This force is always perpendicular to both \(\vec{v}\)and \(\vec{B}\), indicating that it does not change the speed of a particle, only its direction.
Charged Particle Motion
Charged particles moving in magnetic fields exhibit motion influenced by the magnetic force. This force is a result of the cross product of the velocity of the particle and the magnetic field.
  • When a charged particle's velocity is parallel or anti-parallel to the magnetic field, the magnetic force is zero. Therefore, the particle moves in a straight line.
  • When the velocity is perpendicular to the field, the particle undergoes uniform circular motion, continuously changing direction but keeping constant speed.
This behavior results from the perpendicular nature of magnetic force, affecting direction rather than speed. Such phenomena are explained by formulas like the Lorentz force equation: \[\vec{F} = q(\vec{v} \times \vec{B})\]where \(q\)is the charge of the particle.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics establishing a relationship between work done on an object and its change in kinetic energy. Mathematically, it is expressed as:\[W = \Delta KE\]where:
  • \(W\): Work done on the object
  • \(\Delta KE\): Change in kinetic energy
In problems involving magnetic forces, it is crucial to note that the work done by a magnetic force is zero. As the work done is expressed by the dot product:\[W = \vec{F} \cdot \vec{d} = 0\]this zero result originates from the perpendicular nature of the magnetic force to the particle's displacement. Therefore, magnetic forces are unable to change a particle's kinetic energy, emphasizing that velocity direction may change, but speed鈥攁nd consequently, kinetic energy鈥攔emains the same.

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Most popular questions from this chapter

A long straight wire carrying a current of \(30 \mathrm{~A}\) is placed in an external uniform magnetic field of magnitude \(4 \times\) \(10^{-4} \mathrm{~T}\). The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic field in tesla at a point \(2.0 \mathrm{~cm}\) away from the wire is (A) \(10^{-4}\) (B) \(3 \times 10^{-4}\) (C) \(5 \times 10^{-4}\) (D) \(6 \times 10^{-4}\)

A particle of mass \(m\) and charge \(q\) is projected from origin with initial velocity \([u \hat{i}-v \hat{j}]\). Uniform electric and magnetic field exist in the region along \(+y\) direction of magnitudes \(E\) and \(B\), respectively. Then particle will definitively return to the origin if (A) \(\frac{v B}{2 \pi E}\) is an integer (B) \(\frac{\left(u^{2}+v^{2}\right)^{1 / 2}}{[B / \pi E]}\) is an integer (C) \(\frac{v B}{\pi E}\) is an integer (D) \(\frac{v B}{3 \pi E}\) is an integer

The magnetic field due to a current carrying circular loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the centre is \(54 \mu \mathrm{T}\). What will be its value at the centre of loop? (A) \(125 \mu \mathrm{T}\) (B) \(150 \mu \mathrm{T}\) (C) \(250 \mu \mathrm{T}\) (D) \(75 \mu \mathrm{T}\)

The materials suitable for making electromagnets should have [2004] (A) high retentivity and low coercivity. (B) low retentivity and low coercivity. (C) high retentivity and high coercivity. (D) low retentivity and high coercivity.

A magnetic needle is kept in a non-uniform magnetic field. It experiences (A) a torque but not a force. (B) neither a force nor a torque. (C) a force and a torque. (D) a force but not a torque.
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