Problem 9 - Master Resource Book in JEE Main Physics

Question

The magnetic field normal to the plane of a wire of \(n\) turns and radius \(r\) which carries a current \(i\) is measured on the axis of the coil at a small distance \(h\) from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction (a) \((2 / 3) r^{2} / h^{2}\) (b) \((3 / 2) r^{2} / h^{2}\) (c) \((2 / 3) h^{2} / r^{2}\) (d) \((3 / 2) h^{2} / r^{2}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(\frac{3}{2} \frac{h^2}{r^2}\).

Step 1: Understand the Magnetic Field at the Center

The magnetic field at the center of a circular coil with radius \( r \) and \( n \) turns carrying current \( i \) is given by the formula \( B_{center} = \frac{\mu_0 n i}{2r} \), where \( \mu_0 \) is the permeability of free space.

Step 2: Calculate Magnetic Field on the Axis

The magnetic field at a point on the axis at a distance \( h \) from the center of the coil is given by \( B_{axis} = \frac{\mu_0 n i r^2}{2(r^2 + h^2)^{3/2}} \).

Step 3: Calculate the Difference in Magnetic Field

To find the fractional smaller field, note that \( B_{center} - B_{axis} \) needs to be expressed as a fraction of \( B_{center} \). This fraction is what the question asks to determine.

Step 4: Set up the Fraction of the Difference

The fractional difference is \( \frac{B_{center} - B_{axis}}{B_{center}} = 1 - \frac{2r}{\sqrt{(r^2 + h^2)^3}} \approx \frac{3}{2} \cdot \frac{h^2}{r^2} \) for small \( h/r \).

Step 5: Use Approximation for Small Distances

Assume \( h << r \), then \( (r^2 + h^2)^{3/2} \approx r^3 \left(1 + \frac{3h^2}{2r^2}\right) \) and simplify the expression.

Step 6: Compare with Given Options

Compare the derived expression \( \frac{3}{2} \frac{h^2}{r^2} \) to the given options: (a) \( \frac{2}{3} \frac{r^2}{h^2} \), (b) \( \frac{3}{2} \frac{r^2}{h^2} \), (c) \( \frac{2}{3} \frac{h^2}{r^2} \), (d) \( \frac{3}{2} \frac{h^2}{r^2} \).

Step 7: Identify the Correct Option

The fraction \( \frac{3}{2} \frac{h^2}{r^2} \) exactly matches option (d), therefore, this is the correct answer.

Key Concepts

Circular Coil Magnetic Field

The magnetic field within a circular coil is an essential concept in electromagnetism. Let's explore how it works. A circular coil comprises a wire wound in loops or turns, and when current flows through these turns, a magnetic field is created. The principle at play here is that any current-carrying conductor generates a magnetic field around it.

In the case of a circular coil with a certain number of turns, denoted by \( n \), the magnetic field at the center of this coil is intensified due to the cumulative effect of each loop or turn. This magnetic field can be quantified using the formula:

• \( B_{center} = \frac{\mu_0 n i}{2r} \)

where:

• \( \mu_0 \) is the permeability of free space, a constant \( 4\pi \times 10^{-7} \ \text{Tm/A} \)
• \( n \) is the number of turns
• \( i \) is the current
• \( r \) is the radius of the coil

Understanding this formula is crucial, as it forms the basis for calculating the magnetic effects of the coil when the measurement is taken at different points, including along its axis.

Magnetic Field Axis Formula

The magnetic field also exists along the axis of the circular coil, but it varies depending on the distance from the center. If you measure the field at a point along this axis, at a distance \( h \) from the center, the field strength changes. This alteration is captured by the magnetic field axis formula, expressed as:

• \( B_{axis} = \frac{\mu_0 n i r^2}{2(r^2 + h^2)^{3/2}} \)

This formula factors in:

• The distance \( h \) from the coil's center
• How the radius and the distance combine to affect the field strength. The term \((r^2 + h^2)^{3/2}\) reflects the geometrical relationship between different distances.

The denominator gets larger as \( h \) increases, leading to a decrease in the magnetic field strength. Thus, the field on the axis diminishes as you move further away from the coil, unlike the field at the center, demonstrating how spatial distance influences magnetic intensity.

Magnetic Field Calculation

Once you understand how the magnetic field varies from the center to the axis, it becomes possible to calculate and compare these differences. The key question in exercises like these is to determine how the axis field compares to the center field. To solve this, we find the fractional difference:

• Fractional difference: \( \frac{B_{center} - B_{axis}}{B_{center}} \)

When calculated using a specific approximation for small \( h/r \) values, this difference amounts to:

• \( 1 - \frac{2r}{\sqrt{(r^2 + h^2)^3}} \approx \frac{3}{2} \cdot \frac{h^2}{r^2} \)

This calculation illustrates how crucial it is to comprehend geometric and physical approximations. By approximating \( h << r \), which denotes that \( h \) is much smaller than \( r \), simplifications are possible, making the problem manageable and aiding in identifying the correct answer from the given options.