How to Calculate the Force Needed to Move a Conducting Rod

Pull a metal rod to the right across two parallel rails in a magnetic field. With the circuit open, the rod experiences almost no magnetic resistance to its motion. Close the circuit, and suddenly you need to pull harder. The difference is an induced current, which produces a magnetic force opposing the rod’s movement.

So, how much force do you need?

For a conducting rod of length \(L\), moving at constant speed \(v\) perpendicular to a uniform magnetic field of strength \(B\), the additional external force required is

\[ \boxed{F_{\text{pull}}=\frac{B^2L^2v}{R}} \]

where \(R\) is the total resistance of the closed circuit.

This result assumes that the rod, its velocity, and the magnetic field are mutually perpendicular, and that friction and acceleration are negligible.

The important part is how we arrive at this equation. You cannot calculate the magnetic force until you have established the induced EMF and the resulting current.

01Understand the arrangement before calculating

Consider a conducting rod resting across two parallel conducting rails. The rails are connected at one end by a resistor, forming a rectangular circuit. The rod can slide along the rails without losing electrical contact.

A magnet produces a uniform magnetic field perpendicular to the plane of the circuit.

There are three important parts:

• The conducting rod is the moving part of the circuit.
• The rails and resistor complete the electrical loop.
• The magnetic field passes through the area enclosed by the loop.

The loop is not just the rod. It includes both rails and the resistor connecting them.

This matters because the first thing we need to investigate is the change in magnetic flux through the entire loop.

What changes when the rod moves?

Imagine pulling the rod to the right.

The distance between the resistor and the rod increases. Therefore, the area enclosed by the circuit increases.

If the rod has length \(L\), and its distance from the resistor is \(x\), the enclosed area is

\[ A=Lx \]

The magnetic field is constant, but the area is changing.

Consequently, the magnetic flux through the circuit changes.

Think of the loop as a window in steady rain. Making the window wider allows more rain to pass through it, even though the rainfall itself has not changed.

Here, the magnetic field is like the rainfall intensity, while the loop area is like the window area. The analogy stops at the idea of changing area: magnetic flux is a mathematical quantity, not a physical substance flowing through the circuit.

The magnetic flux is given by

\[ \Phi_B=BA\cos\theta \]

where:

• \(\Phi_B\) is magnetic flux, measured in webers (Wb).
• \(B\) is magnetic field strength, measured in teslas (T).
• \(A\) is the enclosed area, measured in square metres (m²).
• \(\theta\) is the angle between the magnetic field and the area normal.

For this arrangement, the field is perpendicular to the circuit, so the magnitude simplifies to

\[ \Phi_B=BA=BLx \]

As the rod moves, \(x\) changes, so the magnetic flux changes.

Predict the direction of the induced current

Suppose the magnetic field points into the page, and the rod moves to the right.

What happens to the magnetic flux?

The area increases, so the magnetic flux directed into the page increases.

According to Lenz’s law, the induced current produces a magnetic field that opposes the change in flux.

Therefore, the induced magnetic field must point out of the page.

Using the right-hand grip rule, the induced conventional current flows anticlockwise around the circuit.

On the moving rod itself, the current flows upwards.

Notice the order of the reasoning:

1. Identify the complete circuit.
2. Determine whether its enclosed area increases or decreases.
3. Determine how the magnetic flux changes.
4. Apply Lenz’s law to find the induced current direction.

Do not guess the current direction simply because the rod moves through a magnetic field.

If the rod moves left instead, the enclosed area decreases. The magnetic flux into the page decreases, so the induced field points into the page. The current reverses and becomes clockwise.

The induced current changes direction, but the magnetic force still opposes the rod’s motion.

02Understand the experiment: EMF is not the same as current

There is an important distinction between induced electromotive force (EMF) and induced current.

An induced EMF can exist even when no current flows.

This is easiest to understand by considering how the sliding-rod experiment can be measured.

The experimental apparatus

Imagine constructing the arrangement using:

• Two parallel conducting rails.
• A conducting rod that slides across the rails.
• A magnet producing a uniform magnetic field through the circuit.
• A resistor connecting the rails.
• An ammeter connected in series to measure current.
• A voltmeter connected across the moving rod.
• A force sensor attached to the rod to measure the pulling force.

The rod is pulled horizontally at a controlled, constant speed.

The experiment can be approached in two stages.

Stage 1: Measure the induced EMF

First, leave the main circuit open and connect a high-resistance voltmeter across the moving rod.

Move the rod through the magnetic field.

The magnetic force on the moving charges inside the rod causes positive and negative charges to separate.

For a rod moving right with the magnetic field directed into the page:

• Positive charges are pushed towards the upper end.
• Negative charges accumulate towards the lower end.
• A potential difference develops between the two ends.

The voltmeter can measure this potential difference.

The important observation is that a voltage can exist even though there is almost no current.

Why?

The voltmeter has a very high resistance, so the current it draws is negligible. There is no substantial closed path for charge to circulate around the main circuit.

This first arrangement measures the induced EMF, but it cannot properly investigate the magnetic braking force produced by a significant induced current.

Stage 2: Close the circuit and measure the force

Now connect the rails through a known resistor and ammeter.

The circuit is complete, so current can flow.

Keep pulling the rod at a constant speed and measure the applied force using the force sensor.

The current in the rod produces a magnetic force opposing its movement.

The force sensor must therefore register a greater pulling force than it did when the circuit was open.

To isolate the magnetic contribution, repeat the measurement with the magnetic field switched off. This provides a baseline for friction and other mechanical resistance.

The difference between the two pulling-force measurements estimates the magnetic drag.

The expected experimental relationships are:

These predictions assume the magnetic field, rod length, and other relevant conditions remain unchanged.

The experiment demonstrates two separate ideas:

• The changing magnetic flux produces an induced EMF.
• When the circuit is closed, that EMF drives a current, which produces a magnetic force on the rod.

This distinction is essential when solving calculation questions.

03Calculate the induced EMF first

Faraday’s law states that the magnitude of the induced EMF is equal to the rate of change of magnetic flux through a circuit.

\[ \varepsilon=\left|\frac{\Delta\Phi_B}{\Delta t}\right| \]

For the sliding-rod arrangement, assume:

• The magnetic field is uniform and constant.
• The rod moves perpendicular to its length.
• The rod moves perpendicular to the magnetic field.
• The other parts of the circuit remain stationary.

The magnetic flux magnitude is

\[ \Phi_B=BLx \]

Since the rod moves at speed \(v\),

\[ v=\frac{\Delta x}{\Delta t} \]

Substituting into Faraday’s law gives

\[ \begin{aligned} \varepsilon &=\frac{BL\Delta x}{\Delta t}\\ &=BLv \end{aligned} \]

Therefore,

\[ \boxed{\varepsilon=BLv} \]

This is the motional EMF produced by the moving rod.

The units are:

• \(B\): tesla (T)
• \(L\): metres (m)
• \(v\): metres per second (m/s)
• \(\varepsilon\): volts (V)

An EMF is energy supplied per unit charge. It is not a mechanical force, despite having the word electromotive in its name.

Now calculate the current

Once you know the induced EMF, you can calculate the current using Ohm’s law.

\[ V=IR \]

For this circuit,

\[ \boxed{I=\frac{BLv}{R_{\text{total}}}} \]

Here, \(R_{\text{total}}\) includes the resistance of the resistor, conducting rod, rails, and any other relevant circuit components.

If the question says the rod and rails have negligible resistance, you can use the resistance of the external resistor alone.

Important: If the circuit is open, the induced EMF can still exist, but the current is approximately zero.

04Calculate the magnetic force on the rod

Once the current is known, you can calculate the magnetic force acting on the conducting rod.

A current-carrying conductor in a magnetic field experiences a force.

The magnitude is

\[ F_B=BIL\sin\theta \]

where:

• \(F_B\) is the magnetic force, measured in newtons (N).
• \(B\) is magnetic field strength, measured in teslas (T).
• \(I\) is current, measured in amperes (A).
• \(L\) is the length of conductor inside the magnetic field, measured in metres (m).
• \(\theta\) is the angle between the current direction and magnetic field.

In our standard arrangement, the current and magnetic field are perpendicular, so \(\sin 90^\circ=1\).

Therefore,

\[ \boxed{F_B=BIL} \]

The magnetic force on the rod acts opposite to its movement.

This follows from Lenz’s law. If the induced current helped the rod move faster, the rod’s motion would produce an induced current that increased the motion producing it. That would contradict the energy transferred into the circuit.

Calculate the external force needed

Suppose you pull the rod to the right at a constant speed.

The magnetic force acts to the left.

For the rod to maintain constant velocity, the resultant horizontal force must be zero.

Therefore,

\[ F_{\text{pull}}=F_B \]

Substitute the magnetic force equation:

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

We already know that

\[ I=\frac{BLv}{R_{\text{total}}} \]

Substituting gives

\[ F_{\text{pull}} =B\left(\frac{BLv}{R_{\text{total}}}\right)L \]

Therefore, the final equation is

\[ \boxed{F_{\text{pull}}= \frac{B^2L^2v}{R_{\text{total}}}} \]

What if the rod is accelerating?

Be careful: the external force does not always equal the magnetic force.

That equality applies when the rod moves at constant speed and other mechanical resistance is negligible.

If the rod is accelerating to the right, Newton’s second law gives

\[ F_{\text{pull}}-F_B-F_{\text{friction}}=ma \]

where:

• \(m\) is the mass of the rod in kilograms.
• \(a\) is its acceleration in metres per second squared.
• \(F_{\text{friction}}\) is any mechanical friction opposing the motion.

If friction is negligible,

\[ F_{\text{pull}}=F_B+ma \]

So always check whether the question specifies constant speed or acceleration.

05Worked examples

06Common mistakes to avoid

These are the mistakes that most often lead to incorrect answers in sliding-rod questions.

One further point: the magnetic force on the current-carrying rod is a field force, not ordinary contact friction. For a refresher on identifying the source of a force, see Contact Forces and Field Forces in HSC Physics.

The charge separation inside an open-circuit rod also creates an electric field. This is related to the electric force acting on individual charges, covered in Force on a Charge in an Electric Field: HSC Physics Guide.

07Questions and solutions

Question 1

A conducting rod of length \(0.30\text{ m}\) moves horizontally to the right at a constant speed of \(4.0\text{ m/s}\) through a uniform magnetic field of \(0.50\text{ T}\), directed into the page.

The rod is part of a closed circuit with a total resistance of \(3.0\,\Omega\).

Calculate:

a) The induced EMF.

b) The induced current.

c) The magnitude and direction of the external force required to maintain constant speed.

Ignore friction.

Solution 1

The required external force is \(0.030\text{ N}\) to the right. The induced EMF is \(0.60\text{ V}\), and the induced current is \(0.20\text{ A}\).

Part a: Induced EMF

Use the motional EMF equation:

\[ \varepsilon=BLv \]

Substitute:

\[ \begin{aligned} \varepsilon &=(0.50)(0.30)(4.0)\\ &=0.60\text{ V} \end{aligned} \]

Part b: Induced current

Use Ohm’s law:

\[ I=\frac{\varepsilon}{R} \]

\[ \begin{aligned} I&=\frac{0.60}{3.0}\\ &=0.20\text{ A} \end{aligned} \]

Part c: External force

The magnetic force magnitude is

\[ F_B=BIL \]

Substituting:

\[ \begin{aligned} F_B &=(0.50)(0.20)(0.30)\\ &=0.030\text{ N} \end{aligned} \]

The rod moves right, so the magnetic force acts left.

The external force must therefore act right to maintain constant speed.

\[ \boxed{F_{\text{pull}}=0.030\text{ N right}} \]

The key is that the external force balances the magnetic force because the rod has zero acceleration.

Question 2

A conducting rod of length \(0.25\text{ m}\) slides along two parallel rails in a uniform magnetic field of \(0.80\text{ T}\), directed into the page.

Initially, the rod is \(1.0\text{ m}\) from the resistor connecting the rails. It moves left at a constant speed of \(2.0\text{ m/s}\) for \(0.25\text{ s}\).

The total circuit resistance is \(2.0\,\Omega\).

Calculate:

a) The change in magnetic flux through the circuit.

b) The average induced EMF.

c) The induced current, including its direction.

d) The magnitude and direction of the magnetic force on the rod.

Solution 2

The magnetic flux decreases by \(0.10\text{ Wb}\), producing an average EMF of \(0.40\text{ V}\). The induced current is \(0.20\text{ A}\) clockwise, and the magnetic force is \(0.040\text{ N}\) to the right.

Part a: Change in magnetic flux

The rod moves left, so the enclosed area decreases.

Calculate the initial and final areas.

Initially,

\[ A_i=Lx_i \]

\[ A_i=(0.25)(1.0)=0.25\text{ m}^2 \]

After \(0.25\text{ s}\), the rod has travelled

\[ \Delta x=vt \]

\[ \Delta x=(2.0)(0.25)=0.50\text{ m} \]

The final separation is

\[ x_f=1.0-0.50=0.50\text{ m} \]

Therefore,

\[ A_f=(0.25)(0.50) =0.125\text{ m}^2 \]

The magnetic flux magnitude is \(\Phi_B=BA\).

Initially,

\[ \Phi_i=(0.80)(0.25) =0.20\text{ Wb} \]

Finally,

\[ \Phi_f=(0.80)(0.125) =0.10\text{ Wb} \]

Taking flux into the page as positive for this calculation,

\[ \begin{aligned} \Delta\Phi_B &=\Phi_f-\Phi_i\\ &=0.10-0.20\\ &=-0.10\text{ Wb} \end{aligned} \]

The negative sign indicates that the flux into the page has decreased.

Part b: Average induced EMF

Using Faraday’s law,

\[ \varepsilon= \left|\frac{\Delta\Phi_B}{\Delta t}\right| \]

\[ \begin{aligned} \varepsilon &=\frac{0.10}{0.25}\\ &=0.40\text{ V} \end{aligned} \]

Part c: Induced current

The magnetic flux into the page decreases.

According to Lenz’s law, the induced field must point into the page to oppose the decrease.

The induced current is therefore clockwise.

Using Ohm’s law,

\[ I=\frac{\varepsilon}{R} \]

\[ \begin{aligned} I&=\frac{0.40}{2.0}\\ &=0.20\text{ A} \end{aligned} \]

Part d: Magnetic force

Use

\[ F_B=BIL \]

\[ \begin{aligned} F_B &=(0.80)(0.20)(0.25)\\ &=0.040\text{ N} \end{aligned} \]

The clockwise current flows downwards through the moving rod.

Using the right-hand rule for the force on a current-carrying conductor, the magnetic force acts to the right.

\[ \boxed{F_B=0.040\text{ N right}} \]

The magnetic force opposes the rod’s leftward movement, as expected from Lenz’s law.

Question 3

A conducting rod of length \(0.50\text{ m}\) moves at a constant speed of \(2.0\text{ m/s}\) through a uniform magnetic field of \(0.60\text{ T}\), directed into the page.

The rod is initially part of an open circuit. A high-resistance voltmeter is connected across its ends.

a) Calculate the induced EMF.

b) Explain why the current in the circuit is approximately zero, despite the induced EMF.

c) The circuit is then closed using a \(3.0\,\Omega\) resistor. Calculate the new current and the magnitude of the magnetic force on the rod.

d) Explain why the magnetic force changes when the switch is closed.

Solution 3

The induced EMF is \(0.60\text{ V}\). With the circuit open, the current and magnetic drag are negligible. After closing the circuit, the current becomes \(0.20\text{ A}\), producing a magnetic force of \(0.060\text{ N}\) opposing the motion.

Part a: Induced EMF

Use

\[ \varepsilon=BLv \]

Substitute:

\[ \begin{aligned} \varepsilon &=(0.60)(0.50)(2.0)\\ &=0.60\text{ V} \end{aligned} \]

Part b: Why is the current approximately zero?

An induced EMF represents energy supplied per unit charge.

It does not guarantee that charge will continuously circulate.

With the main circuit open, there is no complete conducting path around the loop. Charges separate at the ends of the moving rod, producing a potential difference.

The high-resistance voltmeter measures this potential difference while drawing negligible current.

Therefore,

\[ I\approx 0\text{ A} \]

The magnetic force associated with the induced circuit current is also approximately zero.

Part c: Close the circuit

Once the switch is closed, the resistor completes the circuit.

The current is

\[ I=\frac{\varepsilon}{R} \]

\[ \begin{aligned} I&=\frac{0.60}{3.0}\\ &=0.20\text{ A} \end{aligned} \]

Now calculate the magnetic force:

\[ F_B=BIL \]

\[ \begin{aligned} F_B &=(0.60)(0.20)(0.50)\\ &=0.060\text{ N} \end{aligned} \]

The magnetic force is \(0.060\text{ N}\), directed opposite to the rod’s motion.

Part d: Explain the change in magnetic force

The induced EMF exists in both situations because the rod is moving through the magnetic field.

However, closing the circuit allows a substantial current to flow.

The magnetic force on the rod depends on this current:

\[ F_B=BIL \]

With negligible current in the open circuit, the magnetic force is negligible.

With a closed circuit, the current produces a significant magnetic force.

The tempting mistake is to assume that a voltage automatically produces a current. It only does so when there is a conducting path.

Question 4

A conducting rod of length \(0.40\text{ m}\) slides to the right at \(0.50\text{ m/s}\) across two parallel rails.

At a particular instant:

• The rod is \(0.50\text{ m}\) from the resistor connecting the rails.
• The magnetic field has a strength of \(0.80\text{ T}\) and points into the page.
• The magnetic field strength is decreasing at \(0.80\text{ T/s}\).
• The total circuit resistance is \(4.0\,\Omega\).

Assume the field is uniform across the loop.

Calculate the total induced EMF around the circuit, the induced current, and the magnetic force on the rod at that instant.

Explain why using only \(\varepsilon=BLv\) would give an incorrect answer.

Solution 4

The total induced EMF is zero, so the induced current and the magnetic force associated with that current are also zero at this instant.

This is a more subtle application of Faraday’s law because two changes in magnetic flux occur simultaneously.

Step 1: Calculate the enclosed area.

The area is

\[ A=Lx \]

\[ \begin{aligned} A&=(0.40)(0.50)\\ &=0.20\text{ m}^2 \end{aligned} \]

Step 2: Calculate the rate of change of area.

The rod moves to the right, so the area increases.

\[ \frac{dA}{dt}=Lv \]

\[ \begin{aligned} \frac{dA}{dt} &=(0.40)(0.50)\\ &=0.20\text{ m}^2/\text{s} \end{aligned} \]

Step 3: Calculate the total rate of change of magnetic flux.

The magnetic flux directed into the page is

\[ \Phi_{\text{in}}=BA \]

Both \(B\) and \(A\) are changing.

Using the product rule,

\[ \frac{d\Phi_{\text{in}}}{dt} = A\frac{dB}{dt} + B\frac{dA}{dt} \]

Substitute:

\[ \begin{aligned} \frac{d\Phi_{\text{in}}}{dt} &=(0.20)(-0.80)\\ &\quad +(0.80)(0.20)\\ &=-0.16+0.16\\ &=0\text{ Wb/s} \end{aligned} \]

The magnetic field is weakening, but the loop area is increasing.

These two changes exactly cancel at the specified instant.

Step 4: Calculate the total induced EMF.

Using Faraday’s law,

\[ \varepsilon_{\text{total}} = \left|\frac{d\Phi_B}{dt}\right| \]

Therefore,

\[ \boxed{\varepsilon_{\text{total}}=0\text{ V}} \]

Step 5: Calculate the current and magnetic force.

Using Ohm’s law,

\[ I=\frac{0}{4.0}=0\text{ A} \]

The magnetic force due to the induced circuit current is

\[ F_B=BIL \]

\[ F_B=(0.80)(0)(0.40) \]

Therefore,

\[ \boxed{F_B=0\text{ N}} \]

Why does \(BLv\) give the wrong answer on its own?

The motional EMF associated with the moving rod is

\[ \begin{aligned} \varepsilon_{\text{motional}} &=BLv\\ &=(0.80)(0.40)(0.50)\\ &=0.16\text{ V} \end{aligned} \]

However, the decreasing magnetic field also produces an induced EMF around the circuit.

Its magnitude is

\[ \begin{aligned} \varepsilon_{\text{changing field}} &=A\left|\frac{dB}{dt}\right|\\ &=(0.20)(0.80)\\ &=0.16\text{ V} \end{aligned} \]

This contribution acts in the opposite direction to the motional EMF.

The two contributions cancel.

The tempting answer is \(0.16\text{ V}\), obtained by applying the standard motional EMF equation.

The missing consideration is that the magnetic field itself is changing. Faraday’s law applies to the total change in magnetic flux, not just the change caused by the rod’s motion.

Question 5

A conducting rod of length \(0.50\text{ m}\) and mass \(0.20\text{ kg}\) slides without friction along two horizontal conducting rails.

The rails form a closed circuit containing a resistor. The magnetic field is uniform, has a strength of \(0.80\text{ T}\), and points into the page.

A motor pulls the rod horizontally to the right with a constant force of \(0.080\text{ N}\).

Initially, the total circuit resistance is \(2.0\,\Omega\), and the rod has reached a constant terminal speed of \(1.0\text{ m/s}\).

At \(t=0\), an ideal switch changes the total circuit resistance to \(8.0\,\Omega\).

Assume:

• The resistance of the rod and rails is negligible compared with the stated total resistance.
• The motor continues applying the same force.
• Circuit inductance is negligible, so the current adjusts immediately when the resistance changes.
• The magnetic field remains constant.

Determine:

a) The induced EMF immediately after the resistance changes.

b) The induced current and magnetic force immediately after the change.

c) The rod’s initial acceleration after the change.

d) The new terminal speed.

e) The induced current and power dissipated in the resistor at the new terminal speed.

Finally, explain why the current at the new terminal speed is not smaller than its original value, despite the resistance increasing fourfold.

Solution 5

Immediately after the resistance changes, the induced EMF remains \(0.40\text{ V}\), the current falls to \(0.050\text{ A}\), and the magnetic force falls to \(0.020\text{ N}\).

The rod initially accelerates rightwards at \(0.30\text{ m/s}^2\).

Its new terminal speed is \(4.0\text{ m/s}\). At this speed, the current returns to \(0.20\text{ A}\).

The important idea is that increasing resistance changes the rod’s eventual speed. You cannot assume that its speed remains fixed.

Part a: Induced EMF immediately after the resistance changes

The rod has mass, so its velocity cannot change instantaneously.

Immediately after switching, its speed is still

\[ v=1.0\text{ m/s} \]

The magnetic field and rod length are also unchanged.

Therefore,

\[ \varepsilon=BLv \]

\[ \begin{aligned} \varepsilon &=(0.80)(0.50)(1.0)\\ &=0.40\text{ V} \end{aligned} \]

The induced EMF remains

\[ \boxed{\varepsilon=0.40\text{ V}} \]

Part b: Current and magnetic force immediately after switching

The resistance has increased to \(8.0\,\Omega\).

Using Ohm’s law,

\[ I=\frac{\varepsilon}{R} \]

\[ \begin{aligned} I&=\frac{0.40}{8.0}\\ &=0.050\text{ A} \end{aligned} \]

The current has decreased to one-quarter of its original value.

Now calculate the magnetic force:

\[ F_B=BIL \]

\[ \begin{aligned} F_B &=(0.80)(0.050)(0.50)\\ &=0.020\text{ N} \end{aligned} \]

Therefore,

\[ \boxed{F_B=0.020\text{ N left}} \]

The magnetic force acts left, opposing the rod’s rightward motion.

Part c: Initial acceleration

The motor continues pulling with a force of \(0.080\text{ N}\).

The resultant horizontal force is

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

\[ \begin{aligned} F_{\text{net}} &=0.080-0.020\\ &=0.060\text{ N} \end{aligned} \]

Apply Newton’s second law:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} a&=\frac{0.060}{0.20}\\ &=0.30\text{ m/s}^2 \end{aligned} \]

Therefore,

\[ \boxed{a=0.30\text{ m/s}^2\text{ right}} \]

The rod accelerates because the motor’s pulling force is now greater than the magnetic braking force.

Part d: Calculate the new terminal speed

The rod will continue accelerating until the magnetic force increases enough to balance the motor’s pulling force.

At terminal speed,

\[ F_{\text{pull}}=F_B \]

Using

\[ F_B=BIL \]

we obtain

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

Rearrange to calculate the required current:

\[ I=\frac{F_{\text{pull}}}{BL} \]

Substitute:

\[ \begin{aligned} I &=\frac{0.080}{(0.80)(0.50)}\\ &=0.20\text{ A} \end{aligned} \]

The current at the new terminal speed must be \(0.20\text{ A}\).

Now use Ohm’s law to find the required EMF:

\[ \varepsilon=IR \]

\[ \begin{aligned} \varepsilon &=(0.20)(8.0)\\ &=1.60\text{ V} \end{aligned} \]

Since

\[ \varepsilon=BLv \]

rearrange for velocity:

\[ v=\frac{\varepsilon}{BL} \]

Substitute:

\[ \begin{aligned} v &=\frac{1.60}{(0.80)(0.50)}\\ &=4.0\text{ m/s} \end{aligned} \]

Therefore,

\[ \boxed{v_{\text{terminal}}=4.0\text{ m/s}} \]

The new terminal speed is four times the original speed.

Part e: Current and power at the new terminal speed

We already established that the current required for force balance is

\[ \boxed{I=0.20\text{ A}} \]

Calculate the electrical power dissipated in the resistor:

\[ P=I^2R \]

\[ \begin{aligned} P &=(0.20)^2(8.0)\\ &=0.32\text{ W} \end{aligned} \]

Therefore,

\[ \boxed{P=0.32\text{ W}} \]

Why hasn’t the current decreased permanently?

This is the most important conceptual part of the question.

A common prediction is that increasing resistance fourfold must reduce the current to one-quarter of its original value.

That would be correct if the EMF remained constant.

However, the rod does not maintain its original speed. It accelerates because the reduced current produces less magnetic braking.

As the rod speeds up, its induced EMF increases:

\[ \varepsilon=BLv \]

At the new terminal speed, the rod is moving four times faster, so its induced EMF is also four times greater.

The increased EMF exactly compensates for the increased resistance.

Consequently, the final current returns to its original value.

There is another useful check.

At the original terminal speed, the motor supplies mechanical power at the rate

\[ P_{\text{motor}}=F_{\text{pull}}v \]

\[ P_{\text{motor}}=(0.080)(1.0) =0.080\text{ W} \]

At the new terminal speed,

\[ P_{\text{motor}}=(0.080)(4.0) =0.32\text{ W} \]

This matches the electrical power dissipated in the resistor.

The motor supplies four times as much power because the rod now moves four times faster, even though the pulling force is unchanged.

This is a useful application of conservation of energy: the mechanical work supplied by the motor is converted into electrical energy and ultimately dissipated as thermal energy in the resistor.

08Where to go next

The central calculation is now straightforward:

\[ \boxed{ F_{\text{pull}}=\frac{B^2L^2v}{R} } \]

But remember that this is the force needed to maintain constant speed in the idealised arrangement, not a universal expression for every moving conductor.

The deeper connection is between magnetic induction, Lenz’s law, Newton’s second law, and conservation of energy.

Once you can calculate the induced current and magnetic force, you can also investigate how the rod accelerates, how its terminal speed changes, and how mechanical work becomes electrical power.

That is the natural next step when studying electromagnetic induction.