To determine the effect of doubling the diameter of a loop of wire on the induced emf, we need to consider Faraday's Law of Induction, which states that the induced emf (ε) in a loop is proportional to the rate of change of the magnetic flux (Φ) through the loop. The magnetic flux is given by:

$$ \Phi = B \cdot A \cdot \cos(\theta) $$

where:
- $ B $ is the magnetic field strength,
- $ A $ is the area of the loop,
- $ \theta $ is the angle between the magnetic field and the normal to the loop's surface.

The area $ A $ of a circular loop is given by:

$$ A = \pi \left(\frac{d}{2}\right)^2 $$

where $ d $ is the diameter of the loop. If the diameter is doubled, the new diameter $ d' $ is:

$$ d' = 2d $$

The new area $ A' $ is:

$$ A' = \pi \left(\frac{d'}{2}\right)^2 = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 $$

Comparing the new area $ A' $ to the original area $ A $:

$$ A' = 4A $$

Since the induced emf is proportional to the rate of change of the magnetic flux, and the magnetic flux is proportional to the area of the loop, doubling the diameter of the loop results in quadrupling the area. Therefore, the induced emf is also quadrupled.

Thus, the correct answer is:

The induced emf is 4 times as much.

Question

To determine the effect of doubling the diameter of a loop of wire on the induced emf, we need to consider Faraday's Law of Induction, which states that the induced emf (ε) in a loop is proportional to the rate of change of the magnetic flux (Φ) through the loop. The magnetic flux is given by:

$$ \Phi = B \cdot A \cdot \cos(\theta) $$

where:
- $ B $ is the magnetic field strength,
- $ A $ is the area of the loop,
- $ \theta $ is the angle between the magnetic field and the normal to the loop's surface.

The area $ A $ of a circular loop is given by:

$$ A = \pi \left(\frac{d}{2}\right)^2 $$

where $ d $ is the diameter of the loop. If the diameter is doubled, the new diameter $ d' $ is:

$$ d' = 2d $$

The new area $ A' $ is:

$$ A' = \pi \left(\frac{d'}{2}\right)^2 = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 $$

Comparing the new area $ A' $ to the original area $ A $:

$$ A' = 4A $$

Since the induced emf is proportional to the rate of change of the magnetic flux, and the magnetic flux is proportional to the area of the loop, doubling the diameter of the loop results in quadrupling the area. Therefore, the induced emf is also quadrupled.

Thus, the correct answer is:

The induced emf is 4 times as much.

Knowee AI · Accepted Answer