Q.A closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{-4}\ \text{m}^2$ carries a current of $3.0\ \text{A}$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Concept: Magnetic Materials – Magnetization

A current-carrying solenoid behaves like a bar magnet because the magnetic field lines emerge from one end (north pole) and enter the other (south pole), exactly as in a permanent magnet. The sense is determined by the direction of current: using the right-hand rule, if current flows clockwise when viewed from one end, that end is the south pole; anticlockwise gives the north pole.

Reasoning

1. Magnetic moment of a solenoid is $M = N I A$, where $N$ is number of turns, $I$ is current, $A$ is cross-sectional area.
2. Substitute values: $N = 800$, $I = 3.0\ \text{A}$, $A = 2.5 \times 10^{-4}\ \text{m}^2$.
3. Compute: $M = 800 \times 3.0 \times 2.5 \times 10^{-4} = 2400 \times 2.5 \times 10^{-4} = 6000 \times 10^{-4} = 0.6\ \text{A·m}^2$.

The key insight is that a solenoid is simply a long coil of wire. When current flows through it, each turn produces a tiny circular magnetic field. Because the turns are closely wound, these fields add up coherently along the axis, creating a uniform field inside the solenoid that is very similar to the field inside a bar magnet.

Outside the solenoid, the field lines loop from one end to the other — exactly like the field of a bar magnet. One end behaves as a north pole (where field lines emerge) and the other as a south pole (where they re-enter). The sense (which end is north) is given by the right-hand rule: if you curl your fingers in the direction of the current, your thumb points toward the north pole.

The magnetic moment of a solenoid is the total magnetic dipole moment it produces. For a closely wound solenoid, this is simply the product of the number of turns, the current, and the cross-sectional area.

1. Identify the quantities

   Number of turns: $N = 800$

   Current: $I = 3.0\ \text{A}$

   Cross-sectional area: $A = 2.5 \times 10^{-4}\ \text{m}^2$

2. Apply the formula for magnetic moment of a solenoid

   For a single current loop, the magnetic moment is $m = IA$. For $N$ identical turns carrying the same current, the total moment is the vector sum. Since the turns are closely wound and coaxial, their moments add directly:

3. Substitute the values

First multiply $800 \times 3.0 = 2400$

Then $2400 \times 2.5 \times 10^{-4} = 6000 \times 10^{-4} = 0.60$

4. State the result with units The magnetic moment is $0.60\ \text{A·m}^2$.

Method: Magnetic Moment of a Solenoid as an Equivalent Bar Magnet

Concept: A current-carrying solenoid behaves like a bar magnet because the magnetic field lines outside it run from one end (north pole) to the other (south pole), exactly like a bar magnet. The magnetic moment of the solenoid quantifies this behaviour.

Steps

1. Identify the formula The magnetic moment ($\mathbf{m}$) of a closely wound solenoid is given by:

where:

• $N$ = total number of turns
• $I$ = current in amperes
• $A$ = cross-sectional area in $\text{m}^2$

2. List the given values

   • $N = 800$
   • $I = 3.0\ \text{A}$
   • $A = 2.5 \times 10^{-4}\ \text{m}^2$

3. Substitute into the formula

4. Calculate step-by-step

   • First: $800 \times 3.0 = 2400$
   • Then: $2400 \times 2.5 \times 10^{-4} = 6000 \times 10^{-4} = 0.6$

5. Write the final result

Sense in which the solenoid acts like a bar magnet

• The end from which the magnetic field lines emerge (using the right-hand grip rule: curl fingers along current direction, thumb points to north pole) behaves as the north pole.
• The opposite end behaves as the south pole.
• The magnetic moment vector points from the south pole to the north pole inside the solenoid (same direction as the axial field).

Thus, the solenoid is equivalent to a bar magnet of magnetic moment $0.6\ \text{A m}^2$ with its north-south axis along the solenoid's length.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Magnetic Moment with Magnetic Field

The error: Students often try to calculate the magnetic field $B$ inside the solenoid instead of the magnetic moment $m$.

Why it happens: The formula $B = \mu_0 n I$ is drilled in, so students reach for it automatically.

How to avoid:

• Magnetic moment $m$ tells you how strong a magnet the solenoid behaves as — it's a dipole property.
• Magnetic field $B$ is the field produced inside the solenoid.
• For a solenoid acting like a bar magnet, the relevant quantity is:

where $N$ = total turns, $I$ = current, $A$ = cross-sectional area.

Correct calculation:

Mistake 2: Forgetting the Direction (Sense) of the Magnetic Moment

The error: Students give only the magnitude and ignore the "sense" asked in the question.

Why it happens: The phrase "explain the sense" is overlooked — students think only numerical answer is needed.

How to avoid:

• The sense means: which end behaves as North pole and which as South pole.
• Use the right-hand thumb rule:
  • Curl fingers along the direction of current.
  • Thumb points to the North pole of the equivalent bar magnet.
• State clearly:

  The solenoid behaves like a bar magnet with its North pole at the end from which the current appears anticlockwise (when viewed along the axis).

Mistake 3: Using $n$ (turns per unit length) Instead of $N$ (total turns)

The error: Plugging $n = 800/\text{length}$ when length is not given.

Why it happens: Many problems give $n$, so students default to it.

How to avoid:

• Read carefully: "800 turns" means total turns $N = 800$.
• The formula $m = N I A$ uses total turns, not turns per meter.
• If length were given, you'd use $n = N/L$, but here it's not needed.

Mistake 4: Unit Errors in Area Conversion

The error: Writing $2.5 \times 10^{-4}\ \text{m}^2$ as $2.5\ \text{cm}^2$ or forgetting the $10^{-4}$.

Why it happens: Area is given in scientific notation — students misplace the decimal.

How to avoid:

• Always write the unit explicitly: $A = 2.5 \times 10^{-4}\ \text{m}^2$.
• Double-check: $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, so this area is $2.5\ \text{cm}^2$ — but keep it in SI units for calculation.

Mistake 5: Not Explaining the "Bar Magnet" Analogy

The error: Giving only the formula and answer, skipping the conceptual explanation.

Why it happens: Students treat it as a plug-and-chug problem.

How to avoid:

• Write a short paragraph:

  A current-carrying solenoid produces a uniform magnetic field inside, similar to a bar magnet. The field lines emerge from one end (North) and enter the other (South). The magnetic moment points from South to North inside the solenoid, just like in a bar magnet.

Final Answer (Exam-Ready)

Magnetic moment:

Sense:

The solenoid acts like a bar magnet with its North pole at the end where the current flows anticlockwise (viewed along the axis). The magnetic moment vector points from the South pole to the North pole inside the solenoid.