Problem 1868 - JEE Physics

Question

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) \(9 \times 10^{-2}\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: \[B_2 = \frac{\mu_0 I}{2\pi(3r)}\]

Step 3: Find the ratio of B₂ to B₁

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

Step 4: Simplify the ratio

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

Step 5: 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: \[B_2 = B_1 \times \frac{1}{3} = (10^{-8}) \times \frac{1}{3}\]

Step 6: Calculate B₂

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

Step 7: 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) \(3.33 \times 10^{-9}\)

Key Concepts

Magnetic Induction

Magnetic induction is simply a way to measure the strength of a magnetic field at a specific point in space. It tells us how much magnetic influence is present around an object. This measurement is particularly useful for understanding how magnetic forces might affect nearby objects, like a compass needle or electronic circuits.

The formula for finding magnetic induction produced by a long, straight wire carrying a current (I) is:

• \[B = \frac{\mu_0 I}{2\pi r}\]

Here, \(B\) is the magnetic induction, \(\mu_0\) is the permeability of free space, and \(r\) is the distance from the wire. Notice that the magnetic induction is inversely proportional to the distance, meaning it decreases as you move further away from the wire. This shows how magnetic fields weaken with distance. The reasoning becomes essential for calculations like those in the original exercise, where the field needs to be recalculated at different distances.

Ampere's Law

Ampere's Law is an essential principle in magnetism, providing a straightforward way to determine the magnetic field generated by a current-carrying conductor. It states that the line integral of the magnetic field \(B\) along a closed path is proportional to the current \(I\) that passes through the enclosed area. In mathematical terms, Ampere's Law is represented by:

• \[\oint B \cdot dl = \mu_0 I\]

This integral equation means that if you add up all the small magnetic contributions around a loop encircling a wire, it would be directly related to the total current passing through that loop.

For straight wires like in our exercise problem, Ampere's Law assists in deriving the magnetic field formula used. The law simplifies to finding the magnetic field at a point along a path around a current, leading to the direct formula:

• \[B = \frac{\mu_0 I}{2\pi r}\]

This application of Ampere's Law is quite practical, especially when dealing with symmetrical current distributions, helping clearly predict how fields behave.

Permeability of Free Space

Permeability of free space, denoted by \(\mu_0\), represents the ability of a vacuum to support the presence of a magnetic field. It is a fundamental physical constant with a value of approximately \(4\pi \times 10^{-7} \mathrm{~Tm/A}\). This term appears in many formulas relating to magnetic fields and circuits, signifying how vacuum as a medium responds to magnetic influences.

In the magnetic field formula used for our exercise, \(\mu_0\) is crucial:

• \[B = \frac{\mu_0 I}{2\pi r}\]

This equation shows how the permeability of free space influences the strength of the magnetic field produced by a current. It essentially provides a "scaling factor" for the magnetic field strength, considering only the impact of space without any material.
Understanding \(\mu_0\) helps grasp the concept of magnetic fields in environments devoid of material stirring resistance, giving a foundation for calculating fields in much more complex settings.