/*! 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 141 A metal disc of radius \(R=6 \ma... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 141

A metal disc of radius \(R=6 \mathrm{~cm}\) is mounted on a frictionless axle. The current can flow through the axle out along the disc to a sliding contact of rim of the disc. A uniform magnetic field \(B=2 \mathrm{~T}\) is parallel to the axis of the disc. When the current is \(3 \mathrm{~A}\), the disc rotateswith constant angular velocity. The frictional force at the rim between the stationary electrical contact and the rotating rim is \(9 x \times 10^{-2} \mathrm{~N}\). Find the value of \(x\). W

Short Answer

Expert verified
The value of x in the frictional force at the rim is approximately 2.53.

Step by step solution

01

Identify the forces acting on the disc

First, we need to understand the forces acting on the metal disc. There are two forces acting on the rim: 1. The magnetic force (Fm) exerted by the magnetic field on the charges within the rim of the disc. 2. The frictional force (Ff) between the stationary electrical contact and the rotating rim. Since the disc rotates with constant angular velocity, these two forces balance each other, that is, Fm = Ff.
02

Calculate the radius of the disc

As given, the radius of the disc is R = 6 cm. We should convert it to meters for easier calculations. R = 6 cm 脳 (1 m / 100 cm) = 0.06 m
03

Calculate the magnetic force on the charges in the rim

The magnetic force (Fm) can be calculated using the formula: Fm = BIl Where B = magnetic field (2 T), I = current (3 A), and l = length of the conductor in the magnetic field. In this case, the length of the conductor in the magnetic field is the circumference of the disc's rim, given by the formula: l = 2蟺R Substituting the value for R: l = 2蟺(0.06 m) = 0.12蟺 m Now, we can calculate the magnetic force Fm: Fm = (2 T)(3 A)(0.12蟺 m) = 0.72蟺 N
04

Use the balance between magnetic and frictional forces

As mentioned earlier, since the disc rotates with constant angular velocity, the magnetic force (Fm) balances the frictional force (Ff) at the rim. Hence, Ff = Fm = 0.72蟺 N
05

Find the value of x

From the given information, the frictional force at the rim is given by: Ff = 9x 脳 10^{-2} N We can now use the calculated Ff value to find x: 0.72蟺 N = 9x 脳 10^{-2} N Divide both sides by 9 脳 10^{-2} N to find x: x = (0.72蟺 N) / (9 脳 10^{-2} N) 鈮� 2.53 The value of x in the frictional force at the rim is approximately 2.53.

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Key Concepts

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

Rotational Motion
Rotational motion involves objects that spin or rotate around a fixed axis. In this scenario, the metal disc rotates because a magnetic force acts on it. The rotational motion is characterized by constant angular velocity, meaning the speed of rotation doesn鈥檛 change over time. This is crucial because if there were any imbalance in forces, the disc would either speed up or slow down.
A few key points about rotational motion:
  • The axis of rotation is the central point around which the disc spins.
  • Angular velocity remains constant if all torques cancel each other out.
  • In this problem, the forces causing and opposing the motion are perfectly balanced, ensuring steady rotation.
Understanding such concepts helps reinforce how equilibrium plays a role in rotational dynamics.
Frictional Force
Friction is an opposing force that resists motion between two surfaces in contact. Here, the frictional force acts between the metal disc鈥檚 rim and the stationary contact. Even though the axle is frictionless, this contact point must overcome resistance to maintain rotation.
Important aspects of frictional force in this context include:
  • Magnitude: The given frictional force formula is \(9x \times 10^{-2} \mathrm{~N}\), where \(x\) is found by balancing the forces.
  • Balance with Magnetic Force: For steady motion, the frictional resistance equals the magnetic force.
  • Type of Friction: The static or dynamic type determines interaction between the disc and the contact.
Balancing frictional forces is essential not only to maintain rotational motion, but also to ensure efficient functioning of the system without excess wear or unnecessary slowing.
Magnetic Field
A magnetic field exerts force on moving charges, which is fundamental to understanding this problem. Here, a uniform magnetic field of 2 Tesla influences the charges as they flow through the disc. This force causes the disc to rotate by acting perpendicularly to the flow of current.
Key characteristics of a magnetic field in this scenario are:
  • Uniformity: The consistency across the field ensures equal force along all regions of the disc.
  • Magnetic Force Formula: The force is calculated as \(Fm = BIl\), where \(B\) is the field strength, \(I\) is the current, and \(l\) is the effective length of the conductor.
  • Influence on Disc Motion: By providing torque, magnetic fields initiate and sustain the rotation.
Gaining a clear grasp of magnetic fields enhances comprehension of their roles in various applications, like electric motors and generators.

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Most popular questions from this chapter

Three identical bar magnets, each of magnetic moment \(M\), are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is (A) Zero (B) \(3 M\) (C) \(\frac{3 M}{2}\) (D) \(M \sqrt{3}\)

A charged particle enters a region which offers some resistance against its motion, and a uniform magnetic field exists in the region. The particle traces a spiral path as shown. Then (A) angular velocity of particle remains constant. (B) speed of particle decreases continuously. (C) total mechanical energy of the particle remains conserved. (D) net force on the particle is always perpendicular to its direction of motion.

A narrow beam of singly charged carbon ions, moving at a constant velocity of \(6 \times 10^{4} \mathrm{~m} / \mathrm{s}\) is sent perpendicularly in a rectangular region having uniform magnetic field \(B=0.5 \mathrm{~T}\). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\). If mass of an ion \(=A\left(1.66 \times 10^{-27}\right) \mathrm{kg}\), where \(A\) is the mass number then isotopes present in beam are (A) \({ }^{11} \mathrm{C}\) (B) \({ }^{12} \mathrm{C}\) (C) \({ }^{13} \mathrm{C}\) (D) \({ }^{14} \mathrm{C}\)

Two charged particles having charges \(Q\) and \(-Q\) and masses \(m\) and \(4 m\), respectively, enter in uniform magnetic field \(B\) at an angle \(\theta\) with magnetic field from same point with speed \(v\). The displacement from starting point, where they will meet again, is (A) \(\frac{2 \pi m}{Q B} v \sin \theta\) (B) \(\frac{2 \pi m}{Q B} v \cos \theta\) (C) \(\frac{8 \pi m}{Q B} v \cos \theta\) (D) \(\frac{12 \pi m}{Q B} v \cos \theta\)

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]
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