Q.A long straight wire in the horizontal plane carries a current of $50\ \text{A}$ in north to south direction. Give the magnitude and direction of $B$ at a point $2.5\ \text{m}$ east of the wire.

The key idea is the magnetic field due to a long straight current-carrying wire, given by the Biot–Savart law.

Step 1 — Formula

For an infinitely long straight wire, the magnitude of the magnetic field at a perpendicular distance $r$ is

Step 2 — Plug in values

Here $I = 50\ \text{A}$, $r = 2.5\ \text{m}$, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$.

Step 3 — Calculate

Step 4 — Direction

Current flows north to south. Using the right-hand thumb rule (thumb along current, fingers curl in direction of $B$), at a point east of the wire the field points vertically upward (out of the horizontal plane).

Concept and Intuition: Magnetic Force Balance

This problem is a direct application of the Biot-Savart law for an infinitely long straight wire. The key idea is that a current-carrying wire produces a circular magnetic field around it. The strength of this field depends only on the current and the perpendicular distance from the wire — not on the position along the wire. The direction is given by the right-hand thumb rule: if you point your thumb in the direction of the current, your fingers curl in the direction of the magnetic field lines.

Here, the wire is horizontal and carries current from north to south. The point of interest is 2.5 m east of the wire. So we are looking at a point in the horizontal plane, to the east side of the wire. The magnetic field at that point will be perpendicular to both the wire and the line joining the point to the wire — that is, it will be vertical. The question is: up or down?

Let’s work it out step by step.

1. Identify the given data and the formula

   Current, $I = 50\ \text{A}$

   Distance from wire, $r = 2.5\ \text{m}$

   Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$

   For an infinitely long straight wire, the magnitude of the magnetic field at a perpendicular distance $r$ is:

This formula comes directly from Ampere’s circuital law and is valid as long as the wire is long compared to $r$ — which is true here.

2. Calculate the magnitude

   Substitute the values:

Notice that $\pi$ cancels out:

Simplify step by step:

So the magnitude is $4.0 \times 10^{-6}$ tesla.

3. Determine the direction using the right-hand thumb rule

   The current flows from north to south. Point your right-hand thumb in the direction of the current — that is, from north to south. Now, imagine the wire in front of you: north is up, south is down, east is to your right.

   When you point your thumb downward (south), your fingers curl in a specific way. At a point to the east of the wire, your fingers will be pointing upward (out of the plane of the ground). Let’s verify:

   • If the current is going south (downward in a vertical-plane sketch), the magnetic field lines circle clockwise when viewed from above the wire. Note that the wire is horizontal, so “above” means looking down from the ceiling. If current is southward, and you look from above, the field lines circle clockwise. At a point east of the wire, the tangent to the circle is vertical. For clockwise circles, the east-side tangent points upward. So the field is vertically upward.
   Tip

   A quick check: If the current were from south to north (upward), the field at an east point would be downward. Reversing current reverses field direction. Here current is north-to-south, so field at east is upward.

   Thus, the direction is vertically upward (out of the horizontal plane, toward the sky).

4. Common mistake to avoid

   Watch out

   Do not confuse the direction of the magnetic field with the direction of the force on a moving charge. The field itself is circular around the wire; at a given point, it is tangent to that circle. Also, remember that the formula $B = \frac{\mu_0 I}{2\pi r}$ gives the field in tesla only when $r$ is in metres and $I$ in amperes — no unit conversions needed here.

Method: Right-Hand Thumb Rule + Biot–Savart Law for a Long Straight Wire

This is a standard application of the magnetic field due to an infinitely long straight current-carrying wire.

Step 1: Recall the formula

For a long straight wire, the magnitude of 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

Step 2: Substitute the values

Given:

$I = 50\ \text{A}$, $r = 2.5\ \text{m}$

Cancel $\pi$:

Simplify step-by-step:

Step 3: Determine the direction using the Right-Hand Thumb Rule

• Current flows north to south.
• Point is east of the wire.
• Curl the fingers of your right hand in the direction of the magnetic field lines when your thumb points along the current (southward).

At a point east of a southward current, the magnetic field points vertically downward (into the ground).

Final Answer

Magnitude: $\mathbf{4.0 \times 10^{-6}\ T}$

Direction: Vertically downward (into the plane of the horizontal wire)

Here are the most common mistakes students make on this classic magnetic force problem, along with how to avoid each one.

1. Forgetting the Right-Hand Thumb Rule (Direction Error)

The Mistake:

Students often get the direction of the magnetic field wrong — they point it north, south, or even straight up/down incorrectly.

Why it happens:

They either misapply the right-hand thumb rule or forget that the wire is horizontal and the current flows north to south.

How to avoid:

• Visualise the wire: Current flows from north to south.
• Apply the rule: Point your right thumb along the current direction (south). Your curled fingers show the magnetic field direction.
• At a point east of the wire: Your fingers will curl into the page (downward) on the east side.
• Result: The field direction is vertically downward (into the ground).

Key check: For a horizontal wire, the field circles around it — east side gives downward, west side gives upward.

2. Using the Wrong Formula or Missing the Constant

The Mistake:

Using $B = \frac{\mu_0 I}{2\pi r}$ but forgetting the factor of $2$ or using the wrong value of $\mu_0$.

Why it happens:

Rushing or mixing up the formula for a straight wire with that for a solenoid or circular loop.

How to avoid:

• Memorise the exact formula:

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

• Simplify before plugging numbers:

• Always write units — it catches missing factors.

3. Unit Confusion (cm vs m, A vs mA)

The Mistake:

Taking $r = 2.5$ as cm or forgetting to convert km/cm to metres.

Why it happens:

The problem gives $2.5\ \text{m}$ clearly, but in other questions distances are often in cm.

How to avoid:

• Convert everything to SI units before plugging in:
  • Distance → metres
  • Current → amperes
• Check: $2.5\ \text{m}$ is already in metres — no conversion needed here, but always verify.

4. Mixing Up Magnitude and Direction in the Answer

The Mistake:

Giving only the magnitude or only the direction, or stating the direction incorrectly (e.g., “east” or “north”).

Why it happens:

The question explicitly asks for both — students sometimes answer only one part.

How to avoid:

• Read the question twice: “Give the magnitude and direction of $B$.”
• Write your answer in two clear parts:
  • Magnitude: $B = 4 \times 10^{-6}\ \text{T}$ (or $8 \times 10^{-6}\ \text{T}$ — see next point)
  • Direction: Vertically downward (into the ground)

5. Arithmetic Slip in Simplification

The Mistake:

Computing $B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5}$ and getting $4 \times 10^{-6}\ \text{T}$ instead of $8 \times 10^{-6}\ \text{T}$.

Why it happens:

Cancelling $\pi$ incorrectly or dividing $2.5$ into $2 \times 10^{-5}$ wrongly.

How to avoid:

• Cancel step-by-step:

• Double-check with a quick estimate: $\mu_0 I / (2\pi r) \approx (1.26 \times 10^{-6} \times 50) / (6.28 \times 2.5) \approx 63 \times 10^{-6} / 15.7 \approx 4 \times 10^{-6}\ \text{T}$.

Final correct magnitude: $4 \times 10^{-6}\ \text{T}$ (not $8 \times 10^{-6}$).

Quick Summary Checklist

Final Answer (for reference):

• Magnitude: $B = 4 \times 10^{-6}\ \text{T}$
• Direction: Vertically downward (into the ground)