Q.A circular coil of wire consisting of $100$ turns, each of radius $8.0\ \text{cm}$ carries a current of $0.40\ \text{A}$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

The key idea is the magnetic field at the centre of a circular coil, given by the formula for a current-carrying loop.

Step 1 – Write the formula

For a single circular loop of radius $r$ carrying current $I$, the field at the centre is

Step 2 – Adjust for $N$ turns

For $N$ closely wound turns, the fields add, so

Step 3 – Substitute values

$N = 100$, $I = 0.40\ \text{A}$, $r = 8.0\ \text{cm} = 0.080\ \text{m}$, $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$

Simplify:

Why the centre is special

When current flows through a circular loop, each tiny segment of wire produces a magnetic field that points along the axis at the centre. By symmetry, all these contributions add up in the same direction — straight out of the plane of the coil (or into it, depending on current direction). The centre is the simplest point to calculate because every current element is at the same distance $R$ from the point, and the angle between the element and the line joining it to the centre is always $90^\circ$.

For a single turn, the field at the centre is:

where $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$ is the permeability of free space.

If you have $N$ turns closely wound together, each turn contributes the same field at the centre, so the total field is simply $N$ times that of one turn:

This is the formula we’ll use.

Step-by-step calculation

1. Write down the known quantities

   • Number of turns: $N = 100$
   • Current: $I = 0.40\ \text{A}$
   • Radius: $R = 8.0\ \text{cm} = 0.080\ \text{m}$ (always convert to metres)
   • $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$

2. Plug into the formula

3. Simplify step by step

   First, the numerator:

Then multiply by $0.40$:

Denominator:

So:

4. Cancel the factor of $0.16$

   Notice $1.6 / 0.16 = 10$, so:

5. Evaluate numerically

   $\pi \approx 3.1416$, so:

Method: Biot–Savart Law for a Circular Current Loop

This is the standard method for finding the magnetic field at the centre of a current-carrying circular coil.

Steps

1. Recall the formula For a single circular loop of radius $r$ carrying current $I$, the magnetic field at the centre is:

2. Account for multiple turns For a coil with $N$ turns, each turn contributes equally, so:

3. Insert the given values
   • $N = 100$
   • $r = 8.0\ \text{cm} = 0.080\ \text{m}$
   • $I = 0.40\ \text{A}$
   • $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$

4. Simplify step by step

   • Numerator: $4\pi \times 10^{-7} \times 0.40 = 1.6\pi \times 10^{-7}$
   • Denominator: $2 \times 0.080 = 0.16$
   • So: $B = 100 \times \frac{1.6\pi \times 10^{-7}}{0.16}$

   Since $\frac{1.6}{0.16} = 10$, we get:

5. Final magnitude

Numerically:

Key Concept Check

• The field is directly proportional to $N$ and $I$, and inversely proportional to $r$.
• At the centre, all turns produce the field in the same direction (perpendicular to the plane of the coil), so the net field is simply $N$ times that of a single turn.

Here are the most common mistakes students make when solving this standard magnetic field problem, along with how to avoid each.

1. Forgetting the Number of Turns ($N$)

The Mistake:

Students often use the formula for a single loop:

This gives the field for one turn only. For a coil of $N$ turns, the field is $N$ times larger.

How to Avoid:

Always check if the problem mentions “$N$ turns” or “$N$ loops.” The correct formula for a circular coil is:

Here, $N = 100$, so the field is 100 times the field of a single loop.

2. Unit Conversion Errors (cm → m)

The Mistake:

Plugging $R = 8.0$ directly into the formula without converting to metres.

Since $\mu_0$ is in SI units ($4\pi \times 10^{-7}\ \text{T m/A}$), the radius must be in metres.

How to Avoid:

Convert all lengths to metres before substituting:

A quick check: if you forget, your answer will be off by a factor of 100.

3. Using the Wrong Value of $\mu_0$

The Mistake:

Using $\mu_0 = 4\pi \times 10^{-7}$ but forgetting the $4\pi$ in calculations, or using an approximate value like $1.26 \times 10^{-6}$ incorrectly.

How to Avoid:

Write $\mu_0$ explicitly as $4\pi \times 10^{-7}$ and keep it symbolic until the final step. Cancel $\pi$ where possible to simplify arithmetic.

4. Arithmetic Errors with $\pi$ Cancellation

The Mistake:

After substituting, students often multiply and divide without simplifying $\pi$ first, leading to messy decimals and errors.

How to Avoid:

Write the expression step-by-step:

Cancel $\pi$ and simplify numbers first:

Then $160 / 0.16 = 1000$, so:

Final answer: $B = 3.14 \times 10^{-4}\ \text{T}$.

5. Confusing Centre vs. Axis Formula

The Mistake:

Using the formula for field on the axis of a coil (which includes distance $x$ from centre) instead of the simpler centre formula.

How to Avoid:

The problem explicitly says “at the centre.” For the centre of a circular coil:

Only use the axis formula if the point is not at the centre.

6. Forgetting the Direction (When Asked)

The Mistake:

If the problem asks for magnitude and direction, students often give only the magnitude.

How to Avoid:

Use the right-hand thumb rule: curl fingers along current direction, thumb points in direction of $B$ at centre. For a coil, the field is perpendicular to the plane of the coil. If direction is required, state it clearly (e.g., “out of the page” or “along the axis”).

Quick Checklist to Avoid All Mistakes

• ✓ Use $B = \frac{\mu_0 N I}{2R}$ (not single-loop formula)
• ✓ Convert $R$ to metres
• ✓ Keep $\mu_0 = 4\pi \times 10^{-7}$ and cancel $\pi$
• ✓ Simplify numbers step-by-step
• ✓ Use centre formula, not axis formula
• ✓ State direction if asked

Final Answer for Reference: