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A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature \(4 \mathrm{~cm}\). On passing through a metallic sheet it loses half of its kinetic energy. Then, the radius of curvature of the particle is a. \(2 \mathrm{~cm}\) b. \(4 \mathrm{~cm}\) c. \(8 \mathrm{~cm}\) d. \(2 \sqrt{2} \mathrm{~cm}\)

Short Answer

Expert verified
The radius of curvature becomes \(2\sqrt{2} \text{ cm}\).

Step by step solution

01

Understand the Initial Situation

The particle is moving in a uniform magnetic field, which means it follows a circular path due to the Lorentz force. Initially, the radius of its path is given as 4 cm.
02

Effect of Energy Loss on Velocity

When the particle loses half of its kinetic energy, its velocity changes. The kinetic energy is proportional to the square of the velocity (\[ KE \propto v^2 \]), so if the kinetic energy is halved, the velocity becomes \[ v' = \frac{v}{\sqrt{2}} \].
03

Determine the New Radius of Curvature

The radius of curvature \( R \) of a charged particle in a magnetic field is directly proportional to its velocity \( v \), i.e., \( R \propto v \). So after the velocity reduction, the new radius \( R' \) can be mathematically represented as: \[ R' = R \times \frac{v'}{v} = R \times \frac{1}{\sqrt{2}} \].
04

Calculate the New Radius

Substituting the given value of the initial radius \( R = 4 \) cm into the equation, we get: \[ R' = 4 \times \frac{1}{\sqrt{2}} = 4 \div \sqrt{2} = 2\sqrt{2} \text{ cm} \].

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a fundamental concept in physics that relates to the energy of motion. When we talk about kinetic energy, we consider how much energy an object has because it is moving. A crucial part of kinetic energy is that it's highly dependent on the object's velocity. Specifically, the kinetic energy of an object can be calculated using the formula: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass and \( v \) is the velocity of the object. Notice how the velocity is squared, which means even small changes in velocity result in significant changes in kinetic energy. For instance, if the particle's velocity is halved, its kinetic energy is reduced to a quarter of its original value. However, in the given exercise, the particle loses half of its kinetic energy. Thus, its velocity is only reduced by the factor of \( \sqrt{2} \) as the new kinetic energy becomes: \[ KE' = \frac{1}{2} KE \] leading to a new velocity \( v' \) such that: \[ v' = \frac{v}{\sqrt{2}} \]. This relationship is crucial as it helps determine how the particle’s path changes when it passes through a magnetic field.
Lorentz Force and Circular Paths
The Lorentz force is the force acting on a charged particle moving through both electric and magnetic fields. In this exercise, we focus on the magnetic component. This force is key in determining the motion of charged particles in a magnetic field and is given by: \[ F = qvB \] where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength. The Lorentz force always acts perpendicular to the particle's velocity. This is why a charged particle moves in a circular path when in a magnetic field. The central role of the Lorentz force is in changing the direction of the velocity while keeping the particle in circular motion. The particle does not change speed because the force is perpendicular to its motion. Hence, the kinetic energy changes only occur due to external interference, like passing through a metallic sheet as in the exercise. The radius of curvature of this path can be impacted by changes in velocity (as described in the previous section), altering the particle's movement in the magnetic field. This interplay between forces, velocities, and paths demonstrates the fundamental physics orchestrated by the Lorentz force.
The Radius of Curvature and its Dependence
The concept of the radius of curvature is crucial for understanding how charged particles move within a magnetic field. It's defined by the distance from the center of the path's circle to the path itself. The radius of curvature \( R \) is directly related to velocity \( v \) as: \[ R = \frac{mv}{qB} \] Thus, when a charged particle experiences a change in its velocity, the radius of curvature also changes. This is seen clearly when the particle loses part of its kinetic energy and its velocity drops. In our exercise, with an initial radius of \( 4 \text{ cm} \), and the velocity reducing to \( \frac{1}{\sqrt{2}} \) of its original value, the new radius \( R' \) becomes: \[ R' = R \times \frac{1}{\sqrt{2}} = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2} \text{ cm} \]. This illustrates how the physics of magnetic fields and kinetic energy changes affect the motion path of the particles. The radius of curvature helps depict a physical image of the particle's trajectory and how it adjusts due to alterations in kinetic energy and velocity.

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Most popular questions from this chapter

A long cylindrical wire of radius ' \(a\) ' carrles a current \(l\) distributed uniformly over its cross section. If the magnetic fields at distances ' \(r\) and \(R\) from the axis have equal magnitude, then a \(a=\frac{R+r}{2}\) b \(a=\sqrt{R r}\) c. \(a=\operatorname{Rr} / R+r\) d \(a=R^{2} / r\)

A circular curren carrying coil has a radius \(R\). The distance from the center of the coil on the axis where the magnetic induction will be \((1 / 8)^{\mathrm{h}}\) of its value at the center of the coil, is a. \(R / \sqrt{3}\) b \(R \sqrt{3}\) c \(2 R \sqrt{3}\) d \((2 \sqrt{3}) R\)

A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one end. If a rnagnetic field is switched on in the vertical direction, the tension in the string a. will increase b. will decrease c. remains same d. may increase or decrease

Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: \(I_{1}=20 \mathrm{~A}, I_{2}=-6 \mathrm{~A}_{1} I_{3}=12 \mathrm{~A}, I_{4}=-7 \mathrm{~A}, I_{5}=18 \mathrm{~A}\). [Negative currents are opposite in direction to the positive.] The magnetic field induction at a distance of \(10 \mathrm{~cm}\) from the cable is a. \(5 \mu \mathrm{T}\) b. \(15 \mu \mathrm{T}\) c. \(74 \mu \mathrm{T}\) d. \(128 \mathrm{uT}\)

A straight piece of conducting wire with mass \(M\) and length \(L\) is placed on a frictionless incline tilted at an angle \(\theta\) from the horizontal (as shown in Fig. 9.312). There is a uniform, vertical magnetic field at all points (produced by an arrangement of magnets not shown in Fig. 9.312). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to rernain at rest. a. \(\frac{\mathrm{Mg} \tan \theta}{2 L B}\) to the left b. \(\frac{\mathrm{Mg} \tan \theta}{L B}\) to the right c. \(\frac{\mathrm{Mg} \tan \theta}{L B}\) to the left d. \(\frac{3 \mathrm{Mg} \tan \theta}{2 L B}\) to the left

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