The magnetic force acting on a charged particle can never do work because at every instant the force is perpendicular to the velocity. The torque exerted by a magnetic field can do work on a current loop when the loop rotates. Explain how these seemingly contradictory statements can be reconciled

Short Answer

Expert verified

No, the torque doesn’t work in a rotating current loop

Step by step solution

01

What is torque?

Torque is the measurement of the force that can cause an object to rotate about any given axis. It is rotational, force is what causes an object to accelerate in liner motion.

02

Explanation

Now, we know that the magnetic force that the moving charged particle experiences is perpendicular to the velocity, hence the work done is zero. On the other hand, the magnetic field that is acting on the current carrying conductor is perpendicular to the length. Now the torque is perpendicular to the force and the rotation of the loops are also in the same direction. That is why the torque does not work in a rotating current loop

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

Two coils have mutual inductance M=3.25×10-4H. The current in the first coil increases at a uniform rate of 830 A/S. (a) what is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

A 12.4-µF capacitor is connected through a 0.895-MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 20 s

A light bulb glows because it has resistance. The brightness of a light bulb increases with the electrical power dissipated in the bulb. (a) In the circuit shown in Fig. Q25.14a, the two bulbs A and B are identical. Compared to bulb A, does bulb B glow more brightly, just as brightly, or less brightly? Explain your reasoning. (b) Bulb B is removed from the circuit and the circuit is completed as shown in Fig. Q25.14b. Compared to the brightness of bulb A in Fig. Q25.14a, does bulb A now glow more brightly, just as brightly, or less brightly? Explain your reasoning

A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free