How to Analyse a Wire Suspended by Magnetic Force

A straight wire carries current through a magnetic field. Its weight pulls down, but the magnetic force can push it up. Can the magnetic force hold the wire suspended?

Yes, provided the force acts upwards and is large enough to balance the wire’s weight. The method is to calculate the magnetic force on the section of conductor inside the field, then compare it with the gravitational force:

\[
F_B=BIL\sin\theta
\]

For a wire in equilibrium, with the magnetic force acting vertically upwards and no other supporting forces:

\[
F_B=mg
\]

There is an important distinction in this calculation: \(B\) is the magnetic flux density (measured in tesla), whereas \(F_B\) is the magnetic force acting on the wire (measured in newtons). They are not interchangeable.

01Understand the physical arrangement

Imagine a horizontal wire positioned between the poles of a magnet. A current flows through the wire, and part of the conductor lies inside the magnetic field.

Three things matter:

• The magnetic field acts on the current-carrying section of wire.
• The magnetic force acts perpendicular to both the current direction and the magnetic field.
• Gravity acts vertically downwards on the entire wire.

For the magnetic force to suspend the wire, its direction must be upwards.

Consider the arrangement in the diagram:

• Conventional current flows to the right.
• Magnetic field points into the page.
• The magnetic force acts upwards.
• Gravitational force acts downwards.

The upward magnetic force can balance the downward gravitational force.

Notice that only the section of wire inside the magnetic field experiences the magnetic force in this idealised model. However, gravity acts on the entire wire.

This distinction becomes particularly important when a question gives you the total wire length and the length inside the magnetic field separately.

02Distinguish magnetic field strength \(B\) from magnetic force \(F_B\)

The equation for the magnitude of the magnetic force on a straight current-carrying conductor in a uniform magnetic field is:

\[
F_B=BIL\sin\theta
\]

Each quantity represents something different.

The most important distinction is that \(B\) describes the magnetic field, while \(F_B\) describes the force experienced by the conductor.

A stronger magnetic field generally produces a greater magnetic force, provided the current, conductor length, and angle remain unchanged.

Why does the angle matter?

The magnetic force depends on the angle between the direction of conventional current and the magnetic field.

\[
F_B=BIL\sin\theta
\]

There are two useful cases.

Case 1: Conductor perpendicular to the magnetic field

When the current and magnetic field are perpendicular:

\[
\theta=90^\circ
\]

Since \(\sin90^\circ=1\):

\[
F_B=BIL
\]

This gives the maximum magnetic force for a given \(B\), \(I\), and \(L\).

Case 2: Conductor parallel to the magnetic field

When the current and magnetic field are parallel:

\[
\theta=0^\circ
\]

Therefore:

\[
F_B=BIL\sin0^\circ=0
\]

There is no magnetic force on the straight conductor in this arrangement.

For a wire suspended against gravity, the geometry is often arranged so that the current and magnetic field both lie in a horizontal plane. Their magnetic force is then vertical, allowing it to oppose gravity.

Do not confuse the angle between the wire and the magnetic field with the angle between the magnetic force and vertical. The equation uses the former.

03Set the magnetic force equal to the weight

Now connect the magnetic force to Newton’s second law.

The gravitational force acting on a wire of mass \(m\) is:

\[
F_g=mg
\]

where:

• \(m\) is the mass of the wire in kilograms.
• \(g\) is gravitational field strength, approximately \(9.8\,\text{m s}^{-2}\) near Earth’s surface.

If the magnetic force acts vertically upwards, the two forces oppose each other.

Take upwards as the positive direction.

Newton’s second law gives:

\[
\sum F_y=ma_y
\]

Therefore:

\[
F_B-mg=ma_y
\]

For equilibrium, the acceleration is zero:

\[
F_B-mg=0
\]

Hence:

\[
F_B=mg
\]

Substituting the magnetic force equation:

\[
BIL\sin\theta=mg
\]

This is the key relationship for analysing a wire suspended against gravity by magnetic force.

For the common perpendicular arrangement:

\[
\boxed{BIL=mg}
\]

You can rearrange this equation depending on what the question asks you to calculate.

For a non-perpendicular arrangement, include the \(\sin\theta\) factor in the denominator.

What if the wire is not in equilibrium?

Do not automatically set magnetic force equal to weight.

The relationship is:

\[
F_B-mg=ma_y
\]

This gives three possibilities:

If the magnetic force acts downwards, it adds to gravity instead of opposing it.

Also, if a string, support, or other contact force acts on the wire, that force must be included in the net-force calculation. You can only use \(F_B=mg\) when the magnetic force is the only upward supporting force.

04Determine the direction of the magnetic force

Getting the magnitude correct is only half the problem. The magnetic force must also point in the correct direction.

Use Fleming’s left-hand rule:

• First finger: magnetic field direction.
• Second finger: conventional current direction.
• Thumb: magnetic force direction.

A common mistake is to calculate a positive magnetic force and assume that it must act upwards.

The equation \(F_B=BIL\sin\theta\) calculates the magnitude. It does not independently tell you the direction.

What happens if the current is reversed?

Suppose the magnetic force initially acts upwards and balances the wire’s weight.

Now reverse the current without changing the magnetic field.

The magnetic force reverses direction and acts downwards.

If the original magnetic force had magnitude \(mg\), reversing the current produces:

\[
F_{\text{net}}=mg+mg
\]

Therefore:

\[
F_{\text{net}}=2mg
\]

The wire accelerates downwards at:

\[
a=\frac{2mg}{m}=2g
\]

The magnetic force has not disappeared. It now acts in the same direction as gravity.

There is another useful prediction:

• Reversing the current reverses the magnetic force.
• Reversing the magnetic field also reverses the magnetic force.
• Reversing both current and magnetic field leaves the force direction unchanged.

This is why you should establish the direction before substituting numbers.

07A reliable method for HSC questions

When you encounter a question about a wire suspended by magnetic force, work through these decisions before calculating.

A few mistakes are particularly worth avoiding:

• Confusing \(B\) with \(F_B\): Magnetic field strength is measured in tesla, not newtons.
• Using the total wire length: Only the conductor inside the field contributes to the magnetic force in the idealised model.
• Ignoring the angle: Use \(BIL\sin\theta\), unless the conductor is perpendicular to the field.
• Ignoring force direction: An upwards force and a downwards force of equal magnitude have very different consequences.
• Assuming equilibrium: A wire can be accelerating even when it is positioned inside a magnetic field.

The standard equation assumes a straight conductor in a uniform field. Real magnetic fields also have fringe regions near pole edges, where the field changes. HSC problems generally simplify these regions by specifying where the uniform field acts.

08Questions and solutions

The following questions progress from standard calculations to more demanding applications of magnetic force, Newton’s second law, and current direction.

Question 1

A horizontal wire has a mass of \(0.030\,\text{kg}\). A length of \(0.20\,\text{m}\) lies inside a uniform magnetic field.

A current of \(4.0\,\text{A}\) flows through the wire, perpendicular to the magnetic field. The current and field directions produce an upwards magnetic force.

Calculate the magnetic field strength required to keep the wire in equilibrium.

Use \(g=9.8\,\text{m s}^{-2}\).

Solution 1

The required magnetic field strength is \(0.368\,\text{T}\). The magnetic force must equal the wire’s weight because the wire is in equilibrium.

Step 1: Calculate the gravitational force.

\[
F_g=mg
\]

\[
F_g=0.030\times9.8
\]

\[
F_g=0.294\,\text{N}
\]

Therefore:

\[
F_B=0.294\,\text{N}
\]

Step 2: Apply the magnetic force equation.

The conductor is perpendicular to the field:

\[
F_B=BIL
\]

Rearrange:

\[
B=\frac{F_B}{IL}
\]

Step 3: Substitute the values.

\[
B=\frac{0.294}{4.0\times0.20}
\]

\[
B=\frac{0.294}{0.80}
\]

\[
B=0.3675\,\text{T}
\]

Therefore:

\[
\boxed{B=0.368\,\text{T}}
\]

The magnetic force is \(0.294\,\text{N}\) upwards, exactly balancing the gravitational force of \(0.294\,\text{N}\) downwards.

Question 2

A straight horizontal wire has a total length of \(0.50\,\text{m}\) and a mass of \(0.040\,\text{kg}\).

Only \(0.20\,\text{m}\) of the wire is inside a uniform magnetic field of strength \(0.50\,\text{T}\). The wire carries a current of \(3.0\,\text{A}\), perpendicular to the magnetic field.

The magnetic force acts vertically upwards.

Calculate:

(a) The magnitude of the magnetic force acting on the wire.

(b) The initial acceleration of the wire, including its direction.

(c) The current required to keep the wire in equilibrium.

Assume no other vertical forces act on the wire. Use \(g=9.8\,\text{m s}^{-2}\).

Solution 2

The magnetic force is \(0.30\,\text{N}\) upwards, but the wire accelerates downwards at \(2.3\,\text{m s}^{-2}\). A current of \(3.92\,\text{A}\) is required for equilibrium.

Step 1: Calculate the magnetic force.

Only \(0.20\,\text{m}\) of the wire is inside the magnetic field.

Since the conductor is perpendicular to the field:

\[
F_B=BIL
\]

Substitute:

\[
F_B=0.50\times3.0\times0.20
\]

\[
\boxed{F_B=0.30\,\text{N}}
\]

Step 2: Calculate the gravitational force.

The weight acts on the entire wire.

\[
F_g=mg
\]

\[
F_g=0.040\times9.8
\]

\[
F_g=0.392\,\text{N}
\]

Step 3: Determine the net force.

Taking upwards as positive:

\[
F_{\text{net}}=F_B-F_g
\]

\[
F_{\text{net}}=0.30-0.392
\]

\[
F_{\text{net}}=-0.092\,\text{N}
\]

The negative sign indicates that the net force acts downwards.

Step 4: Calculate the acceleration.

Using Newton’s second law:

\[
F_{\text{net}}=ma
\]

Rearrange:

\[
a=\frac{F_{\text{net}}}{m}
\]

\[
a=\frac{-0.092}{0.040}
\]

\[
a=-2.3\,\text{m s}^{-2}
\]

Therefore, the wire accelerates downwards at:

\[
\boxed{2.3\,\text{m s}^{-2}\text{ downwards}}
\]

Step 5: Calculate the current required for equilibrium.

For equilibrium:

\[
F_B=mg
\]

Using the perpendicular conductor equation:

\[
BIL=mg
\]

Rearrange:

\[
I=\frac{mg}{BL}
\]

Substitute:

\[
I=\frac{0.040\times9.8}{0.50\times0.20}
\]

\[
I=\frac{0.392}{0.10}
\]

\[
\boxed{I=3.92\,\text{A}}
\]

The current must increase from \(3.0\,\text{A}\) to \(3.92\,\text{A}\) to provide enough upwards magnetic force to balance the weight.

Question 3

A horizontal wire is initially suspended in equilibrium by a magnetic force.

The wire has an active length of \(0.20\,\text{m}\) inside a uniform magnetic field. The current is \(5.0\,\text{A}\), and the magnetic field strength is \(0.30\,\text{T}\). The conductor is perpendicular to the field.

The magnetic force acts vertically upwards.

The magnetic field is then increased by 20%, while the current is simultaneously decreased by 20%.

(a) Calculate the mass of the wire.

(b) Determine the new magnetic force.

(c) Calculate the wire’s new acceleration, including its direction.

Use \(g=9.8\,\text{m s}^{-2}\).

Solution 3

The wire has a mass of \(0.0306\,\text{kg}\). After the changes, the magnetic force decreases to \(0.288\,\text{N}\), and the wire accelerates downwards at approximately \(0.392\,\text{m s}^{-2}\).

Step 1: Calculate the original magnetic force.

Initially, the wire is in equilibrium, so:

\[
F_B=mg
\]

The magnetic force is:

\[
F_B=BIL
\]

Substitute:

\[
F_B=0.30\times5.0\times0.20
\]

\[
F_B=0.30\,\text{N}
\]

Therefore:

\[
mg=0.30
\]

Step 2: Calculate the mass.

Rearrange:

\[
m=\frac{F_B}{g}
\]

\[
m=\frac{0.30}{9.8}
\]

\[
\boxed{m=0.0306\,\text{kg}}
\]

Step 3: Calculate the new magnetic field and current.

The magnetic field increases by 20%:

\[
B_{\text{new}}=0.30\times1.20
\]

\[
B_{\text{new}}=0.36\,\text{T}
\]

The current decreases by 20%:

\[
I_{\text{new}}=5.0\times0.80
\]

\[
I_{\text{new}}=4.0\,\text{A}
\]

Step 4: Calculate the new magnetic force.

\[
F_{B,\text{new}}=B_{\text{new}}I_{\text{new}}L
\]

\[
F_{B,\text{new}}=0.36\times4.0\times0.20
\]

\[
\boxed{F_{B,\text{new}}=0.288\,\text{N}}
\]

Step 5: Determine the new acceleration.

The gravitational force has not changed:

\[
F_g=0.30\,\text{N}
\]

The net force is:

\[
F_{\text{net}}=0.288-0.30
\]

\[
F_{\text{net}}=-0.012\,\text{N}
\]

Apply Newton’s second law:

\[
a=\frac{F_{\text{net}}}{m}
\]

\[
a=\frac{-0.012}{0.0306}
\]

\[
a=-0.392\,\text{m s}^{-2}
\]

Therefore:

\[
\boxed{a=0.392\,\text{m s}^{-2}
\text{ downwards}}
\]

The interesting part is why the magnetic force decreased.

A tempting prediction is that increasing the field by 20% and decreasing the current by 20% should cancel each other out.

However, magnetic force depends on the product \(BI\), not the sum of the percentage changes.

\[
1.20\times0.80=0.96
\]

The new magnetic force is 96% of its original value. It is therefore 4% smaller, which explains why gravity now produces a net downwards force.

Question 4

A horizontal wire has a mass of \(45\,\text{g}\). A length of \(0.30\,\text{m}\) lies inside a uniform magnetic field.

The wire carries a current of \(6.0\,\text{A}\). The magnetic field lies in the horizontal plane and makes an angle of \(30^\circ\) with the current direction.

The orientation of the current and field produces a vertically upwards magnetic force.

(a) Calculate the magnetic field strength required to suspend the wire in equilibrium.

(b) The magnetic field is then rotated so that it is perpendicular to the wire. The field strength and active wire length remain unchanged. Calculate the new current required for equilibrium.

Use \(g=9.8\,\text{m s}^{-2}\).

Solution 4

The required magnetic field strength is \(0.49\,\text{T}\), and the new equilibrium current is \(3.0\,\text{A}\).

Step 1: Calculate the gravitational force.

Convert the mass:

\[
m=45\,\text{g}=0.045\,\text{kg}
\]

Calculate the weight:

\[
F_g=mg
\]

\[
F_g=0.045\times9.8
\]

\[
F_g=0.441\,\text{N}
\]

Since the wire is in equilibrium:

\[
F_B=0.441\,\text{N}
\]

Step 2: Apply the magnetic force equation.

The current and magnetic field are at \(30^\circ\) to each other.

Therefore:

\[
F_B=BIL\sin30^\circ
\]

Rearrange:

\[
B=\frac{F_B}{IL\sin30^\circ}
\]

Step 3: Substitute the values.

\[
B=\frac{0.441}
{6.0\times0.30\times\sin30^\circ}
\]

Since:

\[
\sin30^\circ=0.5
\]

We obtain:

\[
B=\frac{0.441}{6.0\times0.30\times0.5}
\]

\[
B=\frac{0.441}{0.90}
\]

\[
\boxed{B=0.49\,\text{T}}
\]

Step 4: Calculate the new current when the conductor is perpendicular to the field.

The magnetic field strength remains \(0.49\,\text{T}\).

The new angle is:

\[
\theta=90^\circ
\]

For equilibrium:

\[
BIL=mg
\]

Rearrange:

\[
I=\frac{mg}{BL}
\]

Substitute:

\[
I=\frac{0.441}{0.49\times0.30}
\]

\[
I=\frac{0.441}{0.147}
\]

\[
\boxed{I=3.0\,\text{A}}
\]

The current required has halved.

Why?

The magnetic force depends on \(\sin\theta\). Changing the angle from \(30^\circ\) to \(90^\circ\) doubles the sine factor:

\[
\frac{\sin90^\circ}{\sin30^\circ}
=\frac{1}{0.5}=2
\]

The conductor therefore produces twice the magnetic force for the same current and field strength. Only half the original current is needed to balance gravity.

Question 5

A rectangular conducting loop has a mass of \(0.040\,\text{kg}\). Its horizontal width is \(0.25\,\text{m}\), and its vertical height is \(0.15\,\text{m}\).

The loop carries a current of \(4.0\,\text{A}\) in a clockwise direction, as viewed on the page. The entire loop is positioned inside a uniform magnetic field of strength \(0.40\,\text{T}\), directed into the page.

A student calculates the magnetic force on the top horizontal section:

\[
F_B=BIL
\]

\[
F_B=0.40\times4.0\times0.25
\]

\[
F_B=0.40\,\text{N}
\]

The student notices that the loop’s weight is:

\[
F_g=0.040\times9.8=0.392\,\text{N}
\]

They conclude that the magnetic force is sufficient to suspend the loop.

Is this conclusion correct?

(a) Calculate the magnitude and direction of the magnetic force on each of the four sides of the loop.

(b) Determine the net magnetic force on the complete loop and its initial acceleration if released without any external support.

(c) The magnetic field is now restricted to a narrow horizontal region containing only the top side of the loop. The other three sides are outside the field. Calculate the magnetic field strength required to suspend the loop in this revised arrangement.

Solution 5

The student’s conclusion is incorrect. Although the top section experiences an upwards magnetic force of \(0.40\,\text{N}\), the bottom section experiences an equal downwards force. The magnetic forces on the two vertical sides also cancel horizontally.

The net magnetic force on the complete loop is zero, so the loop cannot be suspended by the uniform field in this arrangement.

Step 1: Calculate the magnetic force on the top horizontal section.

The current flows to the right along the top section, while the magnetic field points into the page.

Using Fleming’s left-hand rule, the force acts upwards.

The magnitude is:

\[
F_{\text{top}}=BIL
\]

\[
F_{\text{top}}=0.40\times4.0\times0.25
\]

\[
\boxed{F_{\text{top}}=0.40\,\text{N upwards}}
\]

Step 2: Calculate the force on the bottom horizontal section.

The bottom section has the same length as the top section.

However, the current flows in the opposite direction.

Therefore, the magnetic force also acts in the opposite direction:

\[
F_{\text{bottom}}=0.40\times4.0\times0.25
\]

\[
\boxed{F_{\text{bottom}}=0.40\,\text{N downwards}}
\]

The upwards and downwards forces cancel.

Step 3: Calculate the forces on the vertical sections.

Each vertical section has a length of \(0.15\,\text{m}\).

The magnetic force magnitude on each is:

\[
F_{\text{side}}=BIL
\]

\[
F_{\text{side}}=0.40\times4.0\times0.15
\]

\[
\boxed{F_{\text{side}}=0.24\,\text{N}}
\]

Using Fleming’s left-hand rule:

• The left vertical section experiences a force to the left.
• The right vertical section experiences a force to the right.

These forces also cancel.

Step 4: Determine the net magnetic force.

The vertical magnetic forces cancel:

\[
F_{\text{vertical,net}}=0.40-0.40=0
\]

The horizontal magnetic forces also cancel:

\[
F_{\text{horizontal,net}}=0.24-0.24=0
\]

Therefore:

\[
\boxed{F_{B,\text{net}}=0\,\text{N}}
\]

The complete loop experiences no net magnetic force.

The student’s mistake was considering only the top section. Current flows continuously around the loop, so every section inside the magnetic field experiences a force.

In a uniform field, the opposite sides experience equal and opposite forces.

Step 5: Calculate the acceleration of the loop.

The only remaining force is gravity.

\[
F_{\text{net}}=mg
\]

Using Newton’s second law:

\[
ma=mg
\]

Therefore:

\[
a=g
\]

The loop accelerates downwards at:

\[
\boxed{a=9.8\,\text{m s}^{-2}
\text{ downwards}}
\]

Step 6: Calculate the required field strength in the revised arrangement.

Now only the top horizontal section lies inside the magnetic field.

The bottom and vertical sections experience no magnetic force in the idealised model.

Consequently, the top section’s upwards magnetic force is no longer cancelled by the other sections.

For equilibrium:

\[
F_B=mg
\]

Using:

\[
BIL=mg
\]

Rearrange:

\[
B=\frac{mg}{IL}
\]

Substitute:

\[
B=\frac{0.040\times9.8}{4.0\times0.25}
\]

\[
B=\frac{0.392}{1.0}
\]

\[
\boxed{B=0.392\,\text{T}}
\]

This is the required magnetic field strength when only the top section experiences the magnetic force.

The important distinction is that the magnetic force on one conductor segment is not necessarily the net magnetic force on the entire circuit. You must account for every current-carrying section inside the field.

09Where this leads next

The central skill in these problems is identifying the actual forces acting on the complete object before deciding which equation to use.

For a suspended wire, you compare magnetic force with weight. For an accelerating wire, you use Newton’s second law. For a conductor with several sections inside a magnetic field, you must consider the direction and magnitude of the force on each section.

The same net-force reasoning appears in gravity as centripetal force in circular orbits, where identifying the actual force and the direction of acceleration is just as important as choosing the correct equation.