Square root of side a is pullout of constant perpendicular magnetic fi...
Understanding the Scenario
In this scenario, a square loop with side length \( a \) is being pulled out of a constant perpendicular magnetic field \( B \) at a constant velocity \( v \). The motion of the loop generates an electromotive force (EMF) due to electromagnetic induction, which according to Faraday's law, is given by the rate of change of magnetic flux through the loop.
Key Concepts
- Electromagnetic Induction
- As the loop is pulled out, the magnetic flux \( \Phi \) through the loop decreases.
- The induced EMF \( \varepsilon \) in the loop can be expressed as:
\[
\varepsilon = -\frac{d\Phi}{dt}
\]
- Resistance and Work Done
- The loop has a resistance \( R \), leading to a current \( I \) flowing through it, given by Ohm's law:
\[
I = \frac{\varepsilon}{R}
\]
- Work Done Against Induced EMF
- The work \( W \) done in pulling the loop out is calculated as:
\[
W = \varepsilon \cdot I \cdot t
\]
- Since the induced EMF is present, the work done is not zero.
Conclusion
In conclusion, the work done in pulling the loop out of the magnetic field is **not zero**. The induced EMF generates a current that opposes the motion (according to Lenz's Law), leading to a resistive force, which means work must be done against this force to maintain the constant velocity \( v \). Therefore, energy is expended in the process, confirming that work done is indeed non-zero.