Problem 1968 - JEE Physics

Question

A small bar magnet has a magnetic moment \(1.2 \mathrm{~A} \cdot \mathrm{m}^{2}\). The magnetic field at a distance \(0.1 \mathrm{~m}\) on its axis will be tesla. (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(2.4 \times 10^{4}\) (d) \(1.2 \times 10^{4}\)

Step-by-Step Solution

Answer

The magnetic field at a distance of \(0.1\,\mathrm{m}\) along the axis of the given small bar magnet is \(2.4 \times 10^{-4} \mathrm{T}\). Answer: (b) \(2.4 \times 10^{-4}\).

Step 1: Recall the formula for magnetic field due to a magnetic dipole

The magnetic field B due to a magnetic dipole at a point on its axis is given by the formula: \[B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}\] Where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (equal to \(4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\)), \(M\) is the magnetic moment, and \(r\) is the distance from the dipole along its axis.

Step 2: Substitute the given values into the formula

Given values are \(M = 1.2 \,\mathrm{A} \cdot \mathrm{m}^2\) and \(r = 0.1\,\mathrm{m}\). Substitute these values into the formula for the magnetic field: \[B =\frac{\mu_0}{4\pi} \cdot \frac{2(1.2\,\mathrm{A} \cdot \mathrm{m}^2)}{(0.1\,\mathrm{m})^3}\]

Step 3: Calculate the magnetic field

Using \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\), let's plug in the value and calculate the magnetic field: \[B = \frac{(4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A})}{4\pi} \cdot \frac{2(1.2\,\mathrm{A} \cdot \mathrm{m}^2)}{(0.1\,\mathrm{m})^3}\] Simplify the expression: \[B = 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A} \cdot \frac{2.4\,\mathrm{A} \cdot \mathrm{m}^2}{0.001\,\mathrm{m}^3}\] \[B = 2.4 \times 10^{-4} \mathrm{T}\]

Step 4: Match the calculated value with the given options

We find that the magnetic field at a distance of \(0.1\,\mathrm{m}\) along the axis of the given small bar magnet is \(2.4 \times 10^{-4} \mathrm{T}\), which corresponds to option (b). Answer: (b) \(2.4 \times 10^{-4}\)

Key Concepts

Magnetic Moment

In the context of magnetism, the magnetic moment is a vector quantity that measures the strength and direction of a magnetic source's magnetism. It represents the extent to which the magnetic source can produce a magnetic field and interact with external magnetic fields. The magnetic moment (M) is usually expressed in units of amperes per square meter (\( ext{A} \, ext{m}^2 \)). A higher magnetic moment indicates a stronger capability to produce a magnetic effect. For example, in our exercise, a bar magnet has a magnetic moment of \( 1.2 \, ext{A} \, ext{m}^2 \), signifying its strength relative to other magnets. This property is essential as it directly influences the strength of the magnetic field a magnet can create at a certain distance.

Magnetic Dipole Formula

The magnetic dipole formula calculates the magnetic field produced by a magnetic dipole, especially at a point on its axis. The formula is expressed as: \[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} \]Here:

• \( B \) is the magnetic field.
• \( \mu_0 \) is the permeability of free space.
• \( M \) is the magnetic moment.
• \( r \) is the distance from the dipole.

This formula shows how the magnetic field decreases rapidly with distance, as demonstrated by the \( r^3 \)term in the denominator.In practical terms, using this formula, we can calculate how strong the magnetic field will be at a particular distance from the dipole, considering the magnetic moment.In this particular exercise, by substituting the given values into the formula, one can determine that the magnetic field at 0.1 meters from the magnet is \( 2.4 \times 10^{-4} \, \text{T} \).

Permeability of Free Space

The permeability of free space, also called the magnetic constant, is a fundamental physical constant used in magnetism calculations. It is denoted by \( \mu_0 \) and defined as \( 4\pi \times 10^{-7} \, \text{T} \cdot \text{m} / \text{A} \).This constant essentially quantifies the extent to which a magnetic field can penetrate the classical vacuum, i.e., space devoid of matter.In the magnetic dipole formula, \( \mu_0 \) plays a crucial role in determining the ultimate value of the magnetic field created by a dipole. Its presence is necessary to convert the units appropriately and achieve the correct formula form.Understanding \( \mu_0 \) is vital as it forms the basis for how magnetic fields operate in free space, providing insight into magnetic force and interactions.