Force on a Current-Carrying Conductor

1.0Understanding Current-Carrying Conductors

A conductor, such as a copper wire, allows free movement of electrons. When a potential difference (voltage) is applied across it, electrons drift, producing an electric current.

• Current is conventionally considered the flow of positive charges, although in reality electrons move in the opposite direction.
• When a conductor carries current, it generates its own magnetic field.

But what happens when such a conductor is placed inside another external magnetic field? That’s where the concept of magnetic force on a current-carrying conductor comes in.

2.0Basics of Magnetic Fields

A magnetic field (B) is a vector field produced by moving charges or magnetic materials. Its strength is measured in Tesla (T).

• A charged particle moving in a magnetic field experiences a force given by the Lorentz force law: $\vec{F}=q(\vec{v}\times \vec{B})$

Here, ( q ) is charge, ( $\vec{v}$ ) is velocity, and ( $\vec{B}$ ) is magnetic field.

For a conductor carrying current, the same principle applies but on a macroscopic level, leading to the derivation of force.

3.0Concept of Magnetic Force on a Conductor

If a conductor of length ( L ) carries current ( I ) and is placed in a magnetic field ( $\vec{B}$ ), it experiences a force due to the interaction between moving charges and the magnetic field.

This force depends on:

• Current in the conductor (( I ))
• Length of conductor in the field (( L ))
• Strength of magnetic field (( B ))
• Angle between conductor and magnetic field (( $\theta$ ))

4.0Derivation of Force on a Current-Carrying Conductor

Current Carrying wire in Uniform Magnetic Field

When a current carrying conductor placed in magnetic field, a magnetic force exerts on each free electron which are present inside the conductor. The resultant of these forces on all the free electrons is called magnetic force on conductor.

Force on a charge element dq moving in MF

$d\vec{F}=dq(\vec{v}\times \vec{B})$

$d\vec{F}=dq\left(\frac{d\vec{\ell }}{dt}\times \vec{B}\right)$

$d\vec{F}=\frac{dq}{dt}(d\vec{\ell }\times \vec{B})$

$d\vec{F}=I(d\vec{\ell }\times \vec{B})$

When current element $Id\vec{\ell }$ is placed in magnetic field $\vec{B}$, it experiences a magnetic force

$d\vec{F}=I(d\vec{\ell }\times \vec{B})$

• Net force on wire due to all such current elements: $\vec{F}=I\int (d\vec{\ell }\times \vec{B})$
• If B is uniform $\vec{F}=I\left(\int d\vec{\ell }\right)\times \vec{B}$
• $\vec{F}=I\left(\int d\vec{\ell }\right)\times \vec{B}$
• $\vec{F}=I({\vec{L}}_{\text{eff}}\times \vec{B})$

$L_{eff}$ is the displacement vector from starting point of current to end point of current.

$L_{eff}$​ is the displacement vector from starting point of current to end point of current.

Current element in a magnetic field does not experience any force if the current in it is parallel or anti–parallel with the field $(\theta =0^{\circ })or(180^{\circ })(d\vec{F}=0)(min.)$

• Current element in a magnetic field experiences maximum force if the current in it is perpendicular with the field $(\theta =90^{\circ })(d\vec{F}=B,I,d\ell )(max.)$
• Magnetic force on current element is always perpendicular to the current element vector and magnetic field vector. $(d\vec{F}\perp I,d\vec{\ell })and(d\vec{F}\perp \vec{B})(always)$

Point of application of magnetic force: On a straight current carrying wire the magnetic force in a uniform magnetic field can be assumed to be acting at its mid point.

This can be used for calculation of torque.

Let us derive step by step:

1. Consider a small length ( $d\vec{l}$ ) of conductor.
2. The current in the conductor is ( I ).
3. In time ( dt ), charge flowing through the segment is: dq = I , dt
4. Each charge moves with drift velocity ($\overrightarrow{v_d}$)
5. The force on each charge due to magnetic field is: $d\vec{F}=dq(\overrightarrow{v_d}\times \vec{B})$
6. Since drift velocity is related to current, after simplification, the total force on conductor becomes: $\vec{F}=I(\vec{L}\times \vec{B})$

Magnitude of force: $F=ILB\sin \theta$

where

• ( I ) = current in the conductor,
• ( L ) = length of conductor inside field,
• ( B ) = magnetic field strength,
• ( $\theta$ ) = angle between conductor and magnetic field.

5.0Direction of Force: Fleming’s Left-Hand Rule

The direction of magnetic force is determined using Fleming’s Left-Hand Rule:

• Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular to each other.
• Forefinger → Direction of magnetic field (( B ))
• Middle finger → Direction of current (( I ))
• Thumb → Direction of force (( F ))

6.0Right Hand Palm Rule

Direction of magnetic force

Magnetic force on the wire we can find by Right Hand Palm Rule

Fingers → In direction of external Magnetic field

Thumb → In direction of L

⊥ to Palm → Direction of magnetic force on wire

Note: For straight wire, direction of L is in the direction of current flow.

7.0Important Cases & Applications

Case 1: Conductor Parallel to Magnetic Field (( $\theta =0^{\circ }$ ))

• ( $\sin 0=0$ )
• Force = 0
• No magnetic force acts on the conductor.

Case 2: Conductor Perpendicular to Magnetic Field (( $\theta =90^{\circ }$ ))

• ( $\sin 90=1$ )
• Maximum force acts: F = I L B

Case 3: Conductor at Angle to Magnetic Field

• General case: $F=ILB\sin \theta$

Case 4: Motion of a Current Loop in a Magnetic Field

• In devices like motors, a rectangular loop of current in a magnetic field experiences torque.
• One side of the loop experiences upward force, the opposite side experiences downward force → causing rotation.

8.0Practical Applications in Daily Life

• Electric motors: Core principle behind converting electrical energy into mechanical motion.
• Loudspeakers: Vibrations created by force on a conductor produce sound.
• Galvanometers: Deflection due to magnetic force measures current.
• Magnetic levitation trains: Based on forces generated by magnetic fields and current.
• Industrial machinery: Cranes use this principle to lift heavy loads.

9.0Common Misconceptions Students Make

• Thinking the force always exists: If the conductor is parallel to the field, force = 0.
• Confusing direction of force: Use Left-Hand Rule, not Right-Hand Rule (which is for generators).
• Assuming magnetic force depends on resistance: It depends only on current, length, magnetic field, and angle.
• Believing electrons are directly pushed: In fact, it’s the overall current-carrying conductor that experiences force.

10.0Related Questions

1. A wire is bent in the form of an equilateral triangle PQR of side 20 cm and carries a current of 2.5 A. It is placed in a magnetic field B of magnitude 2.0 T directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle.

Solution:

Suppose the field and the current have directions as shown in figure. The force on PQ is

$\overrightarrow{F_1}=i\vec{l}\times \vec{B}$

or

$F_1=2.5A\times 20cm\times 2.0T=1.0N$

The rule of vector product shows that the force (F_1) is perpendicular to PQ and is directed towards the inside of the triangle.

The forces ($\overrightarrow{F_2})and(\overrightarrow{F_3}$) on QR and RP can also be obtained similarly.
Both the forces are 1.0 N directed perpendicularly to the respective sides and towards the inside of the triangle.

The three forces ($\overrightarrow{F_1},\overrightarrow{F_2})and(\overrightarrow{F_3}$) will have zero resultant, so that there is no net magnetic force on the triangle. This result can be generalised. Any closed current loop, placed in a homogeneous magnetic field, does not experience a net magnetic force.

2. Figure shows two long metal rails placed horizontally and parallel to each other at a separation y.
A uniform magnetic field B exists in the vertically upward direction. A wire of mass m can slide on the rails. The rails are connected to a constant current source which drives a current i in the circuit. The friction coefficient between the rails and the wire is μ.

(a) What should be the minimum value of μ which can prevent the wire from sliding on the rails?

(b) Describe the motion of the wire if the value of μ is half the value found in the previous part.

3. In the figure shown a semi-circular wire is placed in a uniform B directed toward right. Find the resultant magnetic force and torque on it. 

The wire is equivalent to

∴ θ = 0  ∴ F₍res₎ = 0

Forces on individual parts are marked in the figure by ⊗ and ⊙.

By symmetry their will be pair of forces forming couples.

$\tau ={\int }_0^{\pi /2}i(Rd\theta )B\sin (90-\theta )2R\cos \theta$

$\tau =\frac{i\pi R^2}{2}B$

$\Rightarrow \vec{\tau }=\frac{i\pi R^2}{2}B(-\hat{j})\text{Ans.}$

4. In the figure shown find the resultant magnetic force and torque about “C”, and “P”.

Solution:

$\text{Force on each element is radially outward:}{\tau }_c=0$

Torque About $={\int }_0^{\pi }\left[I(Rd\theta )B\sin 90^0\right]R\sin \theta =2IB{R}^2$