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Chapter 27: Problem 57

If a charged particle of charge \(5 \mu \mathrm{C}\) and mass \(5 \mathrm{~g}\) is moving with constant speed \(5 \mathrm{~m} / \mathrm{s}\) in a uniform magnetic field \(B\) on a curve \(x^{2}+y^{2}=25\), where \(x\) and \(y\) are in metre. The value of magnetic field will be (a) 1 tesla (b) 1 kilo tesla along \(z\) -axis (c) 5 kilo tesla along the \(x\) -axis (d) 1 kilo tesla along any line in the \(x-y\) plane

Short Answer

Expert verified
None of the provided answer choices are correct; the magnetic field should be 0.2 T.

Step by step solution

01

Understand the Particle Motion on a Circle

From the equation of the curve, \(x^2 + y^2 = 25\), it is apparent that the particle is moving along a circle with a radius of \(5\) meters.
02

Determine the Centripetal Force

When a charged particle moves in a magnetic field, the magnetic force acts as a centripetal force. The centripetal force \(F_c\) needed to keep an object of mass \(m\) moving at a speed \(v\) on a circular path of radius \(r\) is given by \(F_c = \frac{m v^2}{r}\).
03

Calculate the Centripetal Force

Use the formula \(F_c = \frac{m v^2}{r}\) with \(m = 0.005 \text{ kg}\), \(v = 5 \text{ m/s}\), and \(r = 5 \text{ m}\):\[F_c = \frac{0.005 \times 5^2}{5} = 0.025 \text{ N}\].
04

Using Lorentz Force Equation

The magnetic force \(F_B\) acting on the charged particle is given by \(F_B = qvB\), where \(q\) is the charge, \(v\) is the speed, and \(B\) is the magnetic field strength.
05

Equating Centripetal and Magnetic Force

Since the magnetic force provides the centripetal force, set \(F_B = F_c\): \[qvB = \frac{m v^2}{r}\].Substitute \(q = 5 \times 10^{-6} \text{ C}\), \(v = 5 \text{ m/s}\), and \(r = 5 \text{ m}\).
06

Solve for the Magnetic Field Strength

Rearrange the equation to solve for \(B\):\[B = \frac{m v}{qr} = \frac{0.005 \times 5}{5 \times 10^{-6} \times 5} = 0.2 \text{ T}\].
07

Convert to Appropriate Units

\(0.2 \text{ T} = 200 \text{ mT}\). The magnetic field value \(0.2 \text{ T}\) or \(200 \text{ mT}\) does not match any provided options exactly, implying the magnetic field's uniform intensity and direction are specified incorrectly in the given choices.

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Key Concepts

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

Centripetal Force
When a charged particle, or any object, moves in a circular path, there is always a force that acts to keep it on that path. This force is called the centripetal force. It always points toward the center of the circle.
The formula to calculate the centripetal force (\(F_c\)) is given by:
  • \(F_c = \frac{m v^2}{r}\)
Where:
  • \(m\) is the mass of the object,
  • \(v\) is the velocity,
  • and \(r\) is the radius of the circular path.
To visualize, imagine swinging a ball tied to a string. The tension in the string provides the centripetal force to keep the ball moving in a circle. In the case of a charged particle in a magnetic field, the magnetic force plays this role, keeping the particle in its path.
Lorentz Force
A charged particle moving in a magnetic field experiences a force called the Lorentz force. This force affects the motion of the particle, especially when it moves perpendicular to the magnetic field.
The Lorentz force (\(F_B\)) is calculated as:
  • \(F_B = qvB\sin胃\)
Where:
  • \(q\) is the charge,
  • \(v\) is the velocity of the particle,
  • \(B\) is the magnetic field strength,
  • and \(胃\) is the angle between the velocity and the magnetic field.
In cases where the particle moves perpendicular to the field, such as in this exercise, \(\sin胃 = 1\). Therefore, the equation simplifies to \(F_B = qvB\). This force is what causes the particle to curve in a path, serving as the centripetal force.
Charged Particle Motion
When a charged particle enters a magnetic field, its path is altered due to the Lorentz force. Instead of continuing in a straight line, the particle is deflected into a curved path. This motion is crucial in various physics applications, like in cyclotrons and particle accelerators.
For a particle moving in a uniform magnetic field:
  • The Lorentz force causes the particle to move in a circular path if it's perpendicular to the field.
  • This force does not work in the traditional sense, as it doesn't change the speed of the particle, only the direction.
In our exercise, the charged particle follows a circular path of radius 5 meters. This motion confirms that the magnetic force acts as the centripetal force necessary for such motion.
Uniform Magnetic Field
A uniform magnetic field is one where the field lines are parallel and equally spaced throughout the region. This type of magnetic field ensures that any charged particle moving through it experiences a constant magnetic force.
Uniform magnetic fields are essential in industrial and scientific applications:
  • They simplify calculations and predictions of particle motions, as forces remain constant.
  • In our exercise, a uniform field is necessary for the charged particle to move in a perfect circle without variability in speed or direction.
Such conditions are frequently used in experimental setups and devices like the mass spectrometer, where precision is key. Understanding the properties of a uniform magnetic field helps predict and control the motion of charged particles effectively.

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Most popular questions from this chapter

If a charged particle is projected in a region of magnetic field, then : (a) the speed of the charged particle continuously changes (b) the magnetic force on the charged particle must be zero (c) the speed of the charged particle remains constant (d) the magnetic force on the charged particle must not be zero

A magnetic needle is kept in non-uniform magnetic field. It experiences: (a) a force and torque (b) a force but not a torque (c) a torque but not a force (d) neither a force nor a torque

A conducting loop carrying a current is placed in a non-uniform magnetic field perpendicular to the plane of loop. Then : (a) loop must experience force (b) loop may experience torque (c) loop must experience torque (d) none of the above

A positive charge is released from the origin at a place where uniform electric field \(E\) and a uniform magnetic field \(B\) exist along the positive \(y\) -axis and positive \(z\) -axis respectively, then: (a) initially the charge particle tends to move along positive \(z\) -axis (b) initially the charged particle tends to move along negative \(y\) -direction (c) initially the charged particle tends to move along positive \(y\) -direction (d) the charged particle moves in \(y-z\) plane

A coil carrying electric current is placed in uniform magnetic field. Then: (a) torque is formed (b) emf is induced (c) both (a) and (b) are correct (d) none of the above.
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