The magnetic flux density at a point distantdfrom a long straight current carrying conductor isB,then its value at distance d/2will beαB.Find the value ofα.Correct answer is '2'. Can you explain this answer?

Question

The magnetic flux density at a point distantdfrom a long straight current carrying conductor isB,then its value at distance d/2will be&alpha;B.Find the value of&alpha;.Correct answer is '2'. Can you explain this answer?

Answer

The magnetic flux density at a point distant d from a long straight current carrying conductor is given by the formula:

B = μ0 * I / (2πd)

where B is the magnetic flux density, μ0 is the permeability of free space, I is the current in the conductor, and d is the distance from the conductor.

If we want to find the value of the magnetic flux density at a distance d/2, we can substitute d/2 for d in the formula:

B' = μ0 * I / (2π * (d/2))

Simplifying this expression:

B' = μ0 * I / (π * d)

Therefore, the value of the magnetic flux density at a distance d/2 is μ0 * I / (π * d).

Answer

The magnetic flux density $ B $ at a distance $ d $ from a long straight current-carrying conductor is given by:

$$
B = \frac{\mu_0 I}{2 \pi d}
$$

where:
- $ \mu_0 $ is the permeability of free space,
- $ I $ is the current in the conductor,
- $ d $ is the distance from the conductor.

If the distance is halved to $ \frac{d}{2} $, the new magnetic flux density $ B' $ becomes:

$$
B' = \frac{\mu_0 I}{2 \pi \left(\frac{d}{2}\right)} = \frac{\mu_0 I}{\pi d} = 2 \times \frac{\mu_0 I}{2 \pi d} = 2B
$$

Thus, the value of $ a $ is $ \boxed{2} $.

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