Q.A $3.0\ \text{cm}$ wire carrying a current of $10\ \text{A}$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27\ \text{T}$. What is the magnetic force on the wire?

The key idea is that a current-carrying conductor in a uniform magnetic field experiences a force given by $F = I L B \sin\theta$, where $\theta$ is the angle between the current direction and the field.

Step 1: The wire is perpendicular to the solenoid’s axis, and the magnetic field inside a solenoid is along its axis. So the angle between the current and the field is $\theta = 90^\circ$, and $\sin 90^\circ = 1$. …

The Core Idea: Magnetic Force on a Current-Carrying Wire

When a current flows through a wire placed in a magnetic field, the moving charges experience a Lorentz force. For a straight wire of length $L$ carrying current $I$ in a uniform magnetic field $B$, the magnitude of the force is:

where $\theta$ is the angle between the direction of the current and the magnetic field vector. The direction of the force is given by Fleming's left-hand rule (or the cross product $\vec{F} = I \vec{L} \times \vec{B}$).

The key insight here is that the solenoid provides a uniform magnetic field along its axis. The wire is placed perpendicular to this axis, meaning the current direction is at $90^\circ$ to the field. That makes $\sin 90^\circ = 1$, so the force is simply $F = I L B$ — no angular complication.

Step-by-Step Solution

1. Identify the given quantities

• Length of wire: $L = 3.0\ \text{cm} = 0.030\ \text{m}$ (always convert to SI units)
• Current: $I = 10\ \text{A}$
• Magnetic field inside solenoid: $B = 0.27\ \text{T}$
• Angle between wire and field: $\theta = 90^\circ$ (since wire is perpendicular to solenoid axis, and the field is along the axis)

2. Write the formula for magnetic force

This is the standard expression derived from the Lorentz force law. The $\sin\theta$ factor accounts for the component of the current that is perpendicular to the field — only that component experiences a force.

3. Substitute the values

Since $\sin 90^\circ = 1$:

4. Calculate step by step …

Method: Magnetic Force on a Current-Carrying Wire (F = BIL sin θ)

This is a direct application of Lorentz force on a current-carrying conductor in a uniform magnetic field.

Steps

1. Identify the formula The magnetic force on a straight current-carrying wire in a uniform field is:

where:

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

2. Extract given data

   • $L = 3.0\ \text{cm} = 0.030\ \text{m}$
   • $I = 10\ \text{A}$
   • $B = 0.27\ \text{T}$
   • Wire is perpendicular to the solenoid's axis → $\theta = 90^\circ$

3. Apply the angle

   $\sin 90^\circ = 1$, so the formula simplifies to:

4. Substitute and calculate

Final Answer …

Here are the common mistakes students make on this Magnetic Force Balance problem, along with how to avoid each.

Mistake 1: Using the wrong formula for magnetic force

Many students mistakenly use the force on a moving charge ($F = qvB$) or the force between two parallel wires, instead of the correct formula for a current-carrying conductor in a uniform field.

How to avoid:

Always identify the situation first. Here, a current-carrying wire is inside a uniform magnetic field (solenoid). The correct formula is:

where:

• $B$ = magnetic field ($0.27\ \text{T}$)
• $I$ = current ($10\ \text{A}$)
• $L$ = length of wire inside the field (not the whole wire length if partially inside)
• $\theta$ = angle between wire and field

Mistake 2: Forgetting to convert cm to metres

A very common slip: using $L = 3.0$ directly without converting to SI units.

How to avoid:

Always convert centimetres to metres before plugging into $F = BIL\sin\theta$.

If you forget, your answer will be 100 times too large.

Mistake 3: Misidentifying the angle $\theta$

Students often assume $\theta = 90^\circ$ is always correct, but sometimes the wire is parallel to the field (force = 0) or at some other angle.

How to avoid:

Read the problem carefully. Here: “perpendicular to its axis” means the wire is perpendicular to the solenoid’s magnetic field. So:

If the wire were parallel, $\sin 0^\circ = 0$ and force would be zero.

Mistake 4: Using the length of the solenoid instead of the wire

Some students take the length of the solenoid (e.g., 30 cm) as $L$, but the force acts only on the wire segment inside the field.

How to avoid:

The problem gives the wire length as $3.0\ \text{cm}$. That is the length inside the solenoid. Use that value (converted to metres).

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