Q.A long straight wire carries a current of $35\ \text{A}$. What is the magnitude of the field $B$ at a point $20\ \text{cm}$ from the wire?

The key idea is the magnetic field around a long straight wire, given by Ampere’s law:

$B = \frac{\mu_0 I}{2\pi r}$.

Step 1: Identify the given values.

Current $I = 35\ \text{A}$, distance $r = 20\ \text{cm} = 0.20\ \text{m}$, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$.

Step 2: Substitute into the formula:

Step 3: Simplify. Cancel $\pi$:

The key to this problem is recognizing that a long straight wire creates a magnetic field that circles around it. The field strength depends only on the current and the perpendicular distance from the wire — not on the length of the wire, as long as the wire is very long compared to the distance. This is a classic application of Ampere’s circuital law, but for a single straight wire, the result is simple enough to use directly.

The direction of the field is tangential to circles centered on the wire (right-hand rule), but the question only asks for magnitude, so we focus on the formula.

1. Identify the given quantities.

   Current $I = 35\ \text{A}$.

   Distance $r = 20\ \text{cm}$. Always convert to SI units: $r = 0.20\ \text{m}$.

2. Write the formula for the magnetic field due to an infinitely long straight wire.

3. Substitute the values.

Notice that $\pi$ cancels out neatly:

4. Simplify step by step. First, $4/2 = 2$, so

Now $35 / 0.20 = 35 \times 5 = 175$, because dividing by 0.20 is the same as multiplying by 5.

So

The magnetic field at a point 20 cm from a wire carrying 35 A is $3.5 \times 10^{-5}$ tesla, which is about the same order as Earth’s magnetic field (roughly $5 \times 10^{-5}\ \text{T}$), so it’s a modest but measurable field.

Method: Biot–Savart Law for a Long Straight Wire

This problem uses the Biot–Savart Law applied to an infinitely long straight current-carrying wire. The law tells us how a current element produces a magnetic field, and when integrated over an infinite wire, it gives a simple formula.

Steps

1. Recall the formula For a long straight wire, the magnetic field at a perpendicular distance $r$ from the wire is:

where:

• $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$ (permeability of free space)
• $I$ = current in amperes
• $r$ = perpendicular distance from the wire in metres

2. Convert units Distance given is $20\ \text{cm}$. Convert to metres:

3. Substitute values

4. Simplify Cancel $\pi$:

Calculate step-by-step:

• Numerator: $4 \times 10^{-7} \times 35 = 140 \times 10^{-7} = 1.4 \times 10^{-5}$
• Denominator: $2 \times 0.20 = 0.40$

So:

5. Final answer

Why this works

The Biot–Savart law shows that each tiny segment of the wire contributes a field that falls off as $1/r^2$, but when summed over an infinite wire, the net field falls off as $1/r$. The direction (given by the right-hand rule) is tangential to circles around the wire — but the question only asks for magnitude.

Here are the most common mistakes students make when solving this standard Biot–Savart / Ampere’s law problem, along with clear ways to avoid each.

1. Forgetting to convert cm to metres

The mistake:

Plugging in $r = 20$ directly into the formula, forgetting that the formula requires SI units (metres).

Why it happens:

The problem gives distance in cm, but the magnetic field constant $\mu_0$ is in $\text{T·m/A}$.

How to avoid:

Always write the conversion step explicitly:

Exam tip: Circle the unit in the question and convert before substituting.

2. Using the wrong formula (field at centre of a loop vs. straight wire)

The mistake:

Using $B = \frac{\mu_0 I}{2R}$ (field at centre of a circular loop) instead of the straight-wire formula.

Why it happens:

Both involve current and distance, so students mix them up under time pressure.

How to avoid:

Memorise the two distinct forms:

• Straight wire:

• Centre of a circular loop:

Mnemonic: Straight wire has a $\pi$ in the denominator; loop does not.

3. Forgetting the factor of $2\pi$ (or using $4\pi$ incorrectly)

The mistake:

Writing $B = \frac{\mu_0 I}{r}$ or $B = \frac{\mu_0 I}{4\pi r}$.

Why it happens:

Students remember $\mu_0 = 4\pi \times 10^{-7}$ and incorrectly cancel the $4\pi$ without the $2\pi$ from the formula.

How to avoid:

Write the formula completely every time:

Then substitute $\mu_0 = 4\pi \times 10^{-7}$:

Now cancel $\pi$ and simplify step by step:

4. Arithmetic errors in simplification

The mistake:

Making a slip when cancelling or multiplying decimals, e.g. $2 \times 0.20 = 0.4$ but then dividing incorrectly.

How to avoid:

Do the arithmetic in two clear stages:

1. Numerator: $4 \times 10^{-7} \times 35 = 140 \times 10^{-7} = 1.4 \times 10^{-5}$
2. Denominator: $2 \times 0.20 = 0.40$

Then:

Final answer: $\boxed{3.5 \times 10^{-5}\ \text{T}}$ (or $35\ \mu\text{T}$)

5. Not stating the direction (when asked)

The mistake:

Giving only magnitude when the question asks for “magnitude of the field $B$” — but in other variants, direction is required.

How to avoid:

If the question ever asks for B (vector), use the right-hand thumb rule:

• Thumb along current direction
• Fingers curl in direction of $B$ (tangent to circles around the wire)

For this question, only magnitude was asked, so direction is not needed — but always check the wording.

Quick checklist before submitting

• Converted cm → m?
• Used $B = \frac{\mu_0 I}{2\pi r}$ (not loop formula)?
• Cancelled $\pi$ correctly?
• Checked arithmetic with powers of 10?
• Included correct unit (T or $\mu$T)?