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Chapter 13: Problem 1868

The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)

Short Answer

Expert verified
The magnetic induction at a distance of 12 cm from the same current-carrying wire is \(3.33 \times 10^{-9}\) tesla.

Step by step solution

01

Write down the formula for the magnetic field of a long current-carrying wire

The formula for the magnetic field B produced by a long current-carrying wire is given by: \[B = \frac{\mu_0 I}{2\pi r}\] where μ₀ is the permeability of free space, I is the current through the wire, and r is the distance from the wire.
02

Calculate the magnetic field at 3 times the distance

We need to find the magnetic induction when the distance r is tripled. Let B₁ be the magnetic induction at a distance r and B₂ be the magnetic induction at a distance 3r. By substituting 3r in place of r, we get: \[B_2 = \frac{\mu_0 I}{2\pi(3r)}\]
03

Find the ratio of B₂ to B₁

Now, we will find the ratio of B₂ to B₁: \[\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi(3r)}}{\frac{\mu_0 I}{2\pi r}}\]
04

Simplify the ratio

Simplify the expression by canceling the common terms in the numerator and denominator: \[\frac{B_2}{B_1} = \frac{r}{3r} = \frac{1}{3}\]
05

Find the new magnetic induction

We know that the magnetic induction at a distance r (4 cm) is given as \(10^{-8}\) tesla. Now, we can find the magnetic induction at a distance 3r (12 cm) by multiplying the known induction by the ratio we found in Step 4: \[B_2 = B_1 \times \frac{1}{3} = (10^{-8}) \times \frac{1}{3}\]
06

Calculate B₂

Calculate the new magnetic induction: \[B_2 = 10^{-8} \times \frac{1}{3} = 3.33 \times 10^{-9}\mathrm{~tesla}\]
07

Choose the correct answer

From our calculations, the magnetic induction at a distance of 12 cm from the same current-carrying wire is \(3.33 \times 10^{-9}\) tesla. Therefore, the correct answer is: (a) \(3.33 \times 10^{-9}\)

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