A circular coi9l of radius r and nuber of turns n cirries a current I. The fields at a small distance h along the axis of the coil and the at the centre of the coil are measured. What is the relation between `B_("centre") and B_("axis")?`

Explanation

To find the relationship between the magnetic field at the center of a circular coil (\(B_{\text{centre}}\)) and the magnetic field at a small distance \(h\) along the axis of the coil (\(B_{\text{axis}}\)), we can follow these steps:

### Step 1: Understand the Magnetic Field at the Center of the Coil
The magnetic field at the center of a circular coil carrying current \(I\) with \(n\) turns and radius \(r\) is given by the formula:
\[ B_{\text{centre}} = \frac{\mu_0 n I}{2r} \]
where \(\mu_0\) is the permeability of free space.

### Step 2: Understand the Magnetic Field along the Axis of the Coil
The magnetic field at a distance \(h\) along the axis of the coil is given by the formula:
\[ B_{\text{axis}} = \frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}} \]

### Step 3: Set Up the Relationship Between \(B_{\text{centre}}\) and \(B_{\text{axis}}\)
To find the relationship, we can take the ratio of \(B_{\text{centre}}\) to \(B_{\text{axis}}\):
\[ \frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 n I}{2r}}{\frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}}} \]

### Step 4: Simplify the Expression
Canceling out the common terms \(\mu_0\), \(n\), \(I\), and \(2\) from the numerator and denominator:
\[ \frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{1/r}{r^2/(r^2 + h^2)^{3/2}} = \frac{(r^2 + h^2)^{3/2}}{r^3} \]

### Step 5: Rearranging the Equation
From the above expression, we can rearrange it to express \(B_{\text{centre}}\) in terms of \(B_{\text{axis}}\):
\[ B_{\text{centre}} = B_{\text{axis}} \cdot \frac{(r^2 + h^2)^{3/2}}{r^3} \]

### Final Relation
Thus, the relationship between \(B_{\text{centre}}\) and \(B_{\text{axis}}\) is:
\[ B_{\text{centre}} = B_{\text{axis}} \cdot \left(1 + \frac{h^2}{r^2}\right)^{3/2} \]