Q.What is the magnitude of magnetic force per unit length on a wire carrying a current of $8\ \text{A}$ and making an angle of $30^\circ$ with the direction of a uniform magnetic field of $0.15\ \text{T}$?

Concept: Magnetic Force Balance — the force on a current-carrying wire in a uniform field depends on the current, field strength, wire length, and the sine of the angle between them.

Step 1: The magnetic force on a straight wire is

$F = I L B \sin\theta$, where $\theta$ is the angle between the current direction and the magnetic field.

Step 2: Per unit length, the force is

$\frac{F}{L} = I B \sin\theta$.

Step 3: Substitute the given values: …

The core idea here is that a magnetic field exerts a force on a moving charge — and since current in a wire is just a stream of moving charges, the wire itself experiences a force. But the force is not simply $I B L$; it depends on the orientation of the wire relative to the field. Only the component of the current perpendicular to the magnetic field contributes to the force. That’s why the sine of the angle appears.

Let’s work through it step by step.

1. Recall the formula for magnetic force on a straight current-carrying wire. When a wire of length $L$ carries a current $I$ in a uniform magnetic field $B$, the magnitude of the magnetic force is:

where $\theta$ is the angle between the direction of the current and the magnetic field vector.

This formula comes directly from the Lorentz force law: $F = q v B \sin\theta$, summed over all moving charges in the wire.

2. We need force per unit length, so divide both sides by $L$:

3. Plug in the given values. Current $I = 8\ \text{A}$, magnetic field $B = 0.15\ \text{T}$, and angle $\theta = 30^\circ$. So:

4. Evaluate $\sin 30^\circ$.

   From trigonometry, $\sin 30^\circ = \frac{1}{2} = 0.5$.

5. Calculate the product: …

Method: Magnetic Force on a Current-Carrying Wire (Biot–Savart / Lorentz Force Law)

We use the Lorentz force on a current element formula:

For a straight wire in a uniform field, the magnitude is:

where:

• $I$ = current in the wire
• $L$ = length of wire
• $B$ = magnetic field strength
• $\theta$ = angle between the wire and the magnetic field

Steps

1. Identify the given quantities

   • $I = 8\ \text{A}$
   • $B = 0.15\ \text{T}$
   • $\theta = 30^\circ$
   • We need force per unit length, so $F/L$ is required.

2. Write the formula for force per unit length

   Divide both sides of $F = I L B \sin\theta$ by $L$:

3. Substitute the values

Recall $\sin 30^\circ = \frac{1}{2}$.

4. Calculate …

Here are the common mistakes students make on this exact Magnetic Force problem, and how to avoid each one.

1. Using the Wrong Formula (Confusing Force on a Charge vs. Wire)

The Mistake:

Students often mix up the formula for force on a moving charge ($F = qvB\sin\theta$) with the formula for force on a current-carrying wire.

How to Avoid:

Always identify the object first:

• Single charge → $F = qvB\sin\theta$
• Current-carrying wire → $F = I L B \sin\theta$

For this problem, you have a wire with current, so use:

2. Forgetting to Divide by Length (Per Unit Length)

The Mistake:

The question asks for force per unit length ($F/L$), but students often calculate the total force $F$ and stop there.

How to Avoid:

Read the question carefully — if it says "per unit length," you need:

Here, $L$ cancels out, so you don't need a length value.

3. Using the Wrong Angle ($\theta$)

The Mistake:

Students sometimes use $30^\circ$ as the angle between the wire and the field, but then plug it into $\cos\theta$ instead of $\sin\theta$, or use the complementary angle ($60^\circ$).

How to Avoid:

The formula uses $\sin\theta$, where $\theta$ is the angle between the current direction and the magnetic field.

Here, the wire makes $30^\circ$ with the field, so $\theta = 30^\circ$ directly.

4. Incorrect Value of $\sin 30^\circ$

The Mistake:

Some students mistakenly use $\sin 30^\circ = \frac{\sqrt{3}}{2}$ (which is $\sin 60^\circ$) or $\frac{1}{\sqrt{2}}$ (which is $\sin 45^\circ$).

How to Avoid:

Memorise the standard angles: