Q.A solenoid of length $0.5\ \text{m}$ has a radius of $1\ \text{cm}$ and is made up of $500$ turns. It carries a current of $5\ \text{A}$. What is the magnitude of the magnetic field inside the solenoid?

The key idea is that for an ideal solenoid (length >> radius), the magnetic field inside is uniform and given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.

Step 1: Find the turn density $n$.

Step 2: Apply the formula for the interior field. The radius is irrelevant here since the field is uniform inside and depends only on $n$ and $I$.

Step 3: Compute the value. …

Why the magnetic field inside a solenoid is so simple

The beauty of a solenoid is that when it's long compared to its radius, the magnetic field inside becomes nearly uniform and parallel to the axis. This isn't a coincidence — it's a direct consequence of symmetry and Ampère's law.

Think of the solenoid as many circular loops stacked side by side. Each loop produces a field along its axis. Near the centre, the contributions from all loops add up constructively, while the field outside nearly cancels. The result: a clean, constant field inside, and almost zero outside.

The key formula comes straight from Ampère's law:

where $n$ is the number of turns per unit length, $I$ is the current, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$ is the permeability of free space.

Let's apply it step by step.

1. Find the number of turns per unit length ($n$) The solenoid has $N = 500$ turns and length $L = 0.5\ \text{m}$.

Notice the radius ($1\ \text{cm}$) is much smaller than the length ($0.5\ \text{m}$). That ratio of 1:50 tells us the solenoid is "long" — so the ideal formula applies with excellent accuracy. If the radius were comparable to the length, we'd need a more complicated calculation.

2. Plug into the formula Current $I = 5\ \text{A}$.

Multiply stepwise:

$1000 \times 5 = 5000$

$4\pi \times 10^{-7} \times 5000 = 4\pi \times 5 \times 10^{-4} = 20\pi \times 10^{-4} = 2\pi \times 10^{-3}$

So $B = 2\pi \times 10^{-3}\ \text{T}$.

3. Numerical value $\pi \approx 3.1416$, so $2\pi \approx 6.2832$. …

Method: Magnetic Field Inside an Ideal Solenoid (Ampere's Circuital Law)

This is a standard application of Ampere's Circuital Law for an ideal (long) solenoid.

Concept (Why this works)

Inside a long solenoid, the magnetic field lines are parallel and uniform. Outside, the field is nearly zero. Ampere's law states:

For a rectangular Amperian loop passing through the solenoid, only the side inside the solenoid contributes to the integral.

Steps

Step 1: Identify the given data

• Length of solenoid, $l = 0.5\ \text{m}$
• Number of turns, $N = 500$
• Current, $I = 5\ \text{A}$
• Radius $= 1\ \text{cm}$ (not needed for an ideal solenoid — field is uniform inside)

Step 2: Find the number of turns per unit length

Step 3: Apply the formula for magnetic field inside an ideal solenoid

where $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$

Step 4: Substitute and calculate …

Here are the common mistakes students make when solving this exact solenoid problem, along with how to avoid each.

1. Using the wrong formula for magnetic field

Mistake:

Students often use the formula for the magnetic field at the centre of a circular loop ($B = \frac{\mu_0 I}{2R}$) or for a straight wire, instead of the solenoid formula.

Why it happens:

The solenoid has circular turns, so it looks like a loop problem. But the field inside a long solenoid is uniform and given by:

where $n$ is the number of turns per unit length.

How to avoid:

• Always identify the geometry first: solenoid → use $B = \mu_0 n I$.
• Remember: $n = \frac{N}{L}$, not just $N$.

2. Confusing $N$ (total turns) with $n$ (turns per metre)

Mistake:

Plugging $N = 500$ directly into $B = \mu_0 N I$, forgetting to divide by length.

Example of error:

How to avoid:

• Write the formula as $B = \mu_0 \frac{N}{L} I$.
• Always compute $n = \frac{N}{L}$ as a separate step.

Correct calculation:

3. Forgetting to convert units (especially length)

Mistake:

Using $L = 0.5$ m correctly but then using radius $1\ \text{cm}$ without converting — or worse, using radius in the formula at all.

Why it happens:

The radius is given, so students think it must be used. But for an ideal solenoid, the field inside is independent of radius (as long as length >> radius).

How to avoid:

• Recognise that radius is a distractor here.
• Only convert units that appear in the formula: $L$ is already in metres, so no conversion needed.
• If radius were needed (e.g., for flux), convert $1\ \text{cm} = 0.01\ \text{m}$.

4. Using $\mu_0$ incorrectly or forgetting it

Mistake:

Using $\mu_0 = 4\pi \times 10^{-7}$ but then dropping the $\pi$ or misplacing powers of 10.

Common slip:

(Forgetting that $1000 \times 5 = 5000$, not 20.)

How to avoid:

• Do arithmetic step-by-step:

5. Forgetting the direction of the field

Mistake: …