Q.A straight wire of mass $200\ \text{g}$ and length $1.5\ \text{m}$ carries a current of $2\ \text{A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$ (Fig. 4.3). What is the magnitude of the magnetic field?

The key idea is Magnetic Force Balance: the upward magnetic force on the wire must exactly cancel its weight for it to float in mid-air.

Step 1 – Forces acting.

The wire’s weight acts downward: $mg$. The magnetic force on a straight current-carrying wire in a uniform field is $F = I l B \sin\theta$. Here the current is horizontal and the field is into the page, so $\theta = 90^\circ$ and $\sin\theta = 1$. The magnetic force is therefore vertical (upward, by Fleming’s left-hand rule).

Step 2 – Balance condition.

For equilibrium:

Step 3 – Solve for $B$.

Step 4 – Substitute values.

Mass $m = 200\ \text{g} = 0.200\ \text{kg}$, $g = 9.8\ \text{m/s}^2$, $I = 2\ \text{A}$, $l = 1.5\ \text{m}$.

The key insight here is that "suspended in mid-air" means the wire is in equilibrium — it isn't accelerating up or down. That tells us the net vertical force on it is zero. The only two vertical forces are its weight (downward) and the magnetic force from the field (upward). So this is a straightforward force balance problem.

Why does the magnetic force point upward? The current is horizontal (to the right, as shown in the figure), and the magnetic field is uniform and directed into the page (indicated by the $\otimes$ symbol). Using the right-hand rule for a current-carrying wire in a magnetic field — point your fingers in the direction of the field (into the page), your thumb in the direction of current (right), and your palm gives the direction of force — you get an upward force. That's exactly what we need to oppose gravity.

Now let's work through the numbers.

1. Write down the known quantities.

   Mass $m = 200\ \text{g} = 0.2\ \text{kg}$ (always convert to SI units).

   Length $l = 1.5\ \text{m}$.

   Current $I = 2\ \text{A}$.

   Acceleration due to gravity $g = 9.8\ \text{m/s}^2$ (standard value unless specified otherwise).

2. Write the force balance equation.

   Magnetic force on a straight wire in a uniform perpendicular field: $F_m = I l B$.

   Weight: $F_g = mg$.

   For equilibrium: $F_m = F_g$, so

3. Solve for $B$.

4. Plug in the numbers.

Method: Magnetic Force Balance (Fleming’s Left Hand Rule + Equilibrium Condition)

Concept

When a current-carrying wire is suspended in mid-air, the magnetic force ($F_m$) acting upward exactly balances the gravitational force ($F_g$) acting downward. This is a simple equilibrium problem.

Steps

Step 1 – Identify the forces

• Downward force: weight of the wire

• Upward force: magnetic force on a straight current-carrying conductor in a uniform perpendicular field

Step 2 – Apply equilibrium condition

For the wire to be suspended (no acceleration):

Step 3 – Solve for $B$

Step 4 – Substitute values (with correct units)

• $m = 200\ \text{g} = 0.200\ \text{kg}$
• $g = 9.8\ \text{m/s}^2$ (standard value)
• $I = 2\ \text{A}$
• $L = 1.5\ \text{m}$

Step 5 – Calculate

Key Exam Point

• The magnetic field must be horizontal and perpendicular to the wire for maximum force ($\theta = 90^\circ$, so $\sin\theta = 1$).
• Always convert mass to kg (not grams) before using $F = mg$.

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to Convert Mass to Kilograms

The error: Students take $m = 200\ \text{g}$ directly into $F = mg$, getting $F = 200 \times 9.8 = 1960\ \text{N}$ — a wildly wrong answer.

Why it happens: The formula $F = mg$ works in SI units, but mass must be in kilograms. Grams are not SI.

How to avoid: Always check units before plugging into any formula. Write:

Then:

Mistake 2: Using the Wrong Force Balance Equation

The error: Setting $BIL = mg$ without checking directions, or forgetting that the wire is suspended in mid-air — meaning magnetic force upward balances gravitational force downward.

Why it happens: Students rush to write $F_B = F_g$ without visualizing the free-body diagram.

How to avoid: Draw a quick sketch:

• Weight ($mg$) acts downward
• Magnetic force ($BIL$) acts upward (using Fleming's left-hand rule)
• For equilibrium: $BIL = mg$ (magnitudes equal, directions opposite)

Mistake 3: Confusing Length and Current Direction

The error: Using $L = 1.5\ \text{m}$ correctly but then misapplying Fleming's left-hand rule — thinking the field must be vertical or horizontal without checking.

Why it happens: The problem says "uniform horizontal magnetic field." Students sometimes assume the field is vertical because "magnetic field" often points north-south in examples.

How to avoid: Read carefully: horizontal field means $B$ is perpendicular to the wire (which is straight and presumably horizontal too). For maximum force, $B$ must be perpendicular to both current and wire length. The given arrangement ensures this.

Mistake 4: Arithmetic Errors in Final Calculation

The error: Computing $B = \frac{mg}{IL}$ incorrectly — e.g., $B = \frac{1.96}{2 \times 1.5} = \frac{1.96}{3} \approx 0.653$ instead of $0.653\ \text{T}$.

Why it happens: Rushing through division or forgetting units.

How to avoid: Write step-by-step:

Double-check: $0.653\ \text{T}$ is a reasonable value for a lab magnet — not too large, not too small.

Mistake 5: Not Stating the Final Answer with Units

The error: Writing $B = 0.653$ without "T" (tesla).

Why it happens: In exams, students forget that units carry marks.

How to avoid: Always write the unit after every numerical answer. The correct final answer is:

Quick Checklist to Avoid All Mistakes