A spatially uniform time dependent magnetic field is changing with time at a constant rate of one tesla per second a unit positive charge is move around the circle of radius R is equals to 2 m perpendicular to this field the magnitude of the work done on charge for one complete revolution is?

Question

Answer

Understanding the Scenario
In this scenario, a unit positive charge is moving in a circular path of radius $ R = 2 \, \text{m} $ perpendicular to a time-dependent magnetic field that changes at a rate of $ 1 \, \text{T/s} $.

Magnetic Field and Charge Interaction
- A changing magnetic field induces an electric field according to Faraday's law of electromagnetic induction.
- The induced electric field $ E $ can be determined using the equation:
$$
E = -\frac{d\Phi_B}{dt}
$$
- Here, $ \Phi_B $ is the magnetic flux given by $ \Phi_B = B \cdot A $, where $ A $ is the area enclosed by the circular path.

Calculating the Induced Electric Field
- The area $ A $ of the circle is calculated as:
$$
A = \pi R^2 = \pi (2)^2 = 4\pi \, \text{m}^2
$$
- Since the magnetic field changes at $ 1 \, \text{T/s} $, the induced electric field becomes:
$$
E = 1 \, \text{T/s} \cdot A = 1 \cdot 4\pi = 4\pi \, \text{V/m}
$$

Work Done on the Charge
- The work done $ W $ on a charge $ q $ moving in an electric field $ E $ over a distance $ d $ is given by:
$$
W = qEd
$$
- For one complete revolution, $ d $ is equal to the circumference of the circle:
$$
d = 2\pi R = 2\pi \cdot 2 = 4\pi \, \text{m}
$$
- Substituting $ q = 1 \, \text{C} $, $ E = 4\pi \, \text{V/m} $, and $ d = 4\pi \, \text{m} $:
$$
W = 1 \cdot 4\pi \cdot 4\pi = 16\pi^2 \, \text{J}
$$

Conclusion
- The magnitude of the work done on the unit positive charge for one complete revolution is $ 16\pi^2 \, \text{J} $.

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams