experiment to show force on a current carrying conductor

Force on A Current-carrying Conductor & Fleming’s Left Hand Rule

Properties Of Magnets
Induced Magnetism & Electrical Method Of Magnetisation
Magnetic Field & Magnetic Field Lines
Temporary & Permanent Magnets
Magnetic Field Due To Current In A Straight Wire
Magnetic Field Due To Current In A Solenoid
Electric Bell
Circuit Breaker
Force On A Current-carrying Conductor & Fleming's Left Hand Rule
Electromagnetic Induction & Lenz's Law
A.C. Generator
Workings Of A Transformer
Practice MCQs For Simple Phenomena Of Magnetism, Force on Conductor in a Magnetic Field, Electromagnetic Force
O Level Physics Topic List
Magnetic Flux Density
Motion of a moving charge in a uniform magnetic field
Velocity Selector
Magnetic Fields due to currents

Force on a current-carrying conductor.

When current-carrying conductor is placed in a magnetic field, it will experience a force when the magnetic field direction is not parallel to the current direction. The magnitude of the force is maximum when the magnetic field and current directions are mutually perpendicular to each other. The force decreases when the angle between the magnetic field and current directions is smaller than $90^{\circ}$.

Factors Affecting Magnetic Force On A Current-carrying Conductor In A Magnetic Field:

Angle between the magnetic field and current directions (More about this below)
Magnetic field strength (Stronger magnetic field $\rightarrow$ stronger force)
Amount of current in conductor (Higher current $\rightarrow$ stronger force)
Length of conductor within magnetic field (Longer conductor $\rightarrow$ stronger force)

If the current direction is PARALLEL to the magnetic field, there will NO force on the conductor by the magnetic field. The magnitude of the force is MAXIMUM when the angle between the magnetic field and current direction is $90^{\circ}$.

This is commonly exploited to produce a turning effect in a current-carrying coil to produce an electric motor.

It does not have to be a current carrying conductor to experience a force due to the magnetic field. The magnetic field actually interacts with the moving electrons in the conductor to produce the force. Hence, electrons that are moving in the direction perpendicular to the magnetic field will experience the force as well. This means that if you pass an electron beam through a magnetic field, it will be deflected. (provided it is perpendicular)

Fleming’s Left-hand Rule – Direction Of The Force

When a conductor carrying a current is placed in a magnetic field, the conductor experiences a magnetic force.

The direction of this force is always right angles to the plane containing both the conductor and the magnetic field, and is predicted by Fleming’s Left-Hand Rule.
F is Force, B is Magnetic field, I is current.
From the name of the rule, use your left hand.

E.g. If current flowing towards to right and the magnetic field is pointing into the paper, the direction of the force is predicted by the Fleming’s left hand rule to be upwards.

Formula For Magnetic Force On A Current-carrying Conductor In A Magnetic Field (A Level)

Referring to the diagram above, F is Force, B is Magnetic field, I is current.

$F = B I l \, sin \, \theta$, where

F is force acting on a current carrying conductor,B is magnetic flux density (magnetic field strength),
I is magnitude of current flowing through the conductor,
$l$ is length of conductor,
$\theta$ is angle that conductor makes with the magnetic field.

When the conductor is perpendicular to the magnetic field, the force will be maximum. When it is parallel to the magnetic field, the force will be zero.

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Bridge Course Class 12th Physics

Course: bridge course class 12th physics > unit 3.

Magnetic field due to a solenoid
Magnetic field due to a current-carrying solenoid

Force on a current-carrying conductor in a magnetic field

Force on a Current-Carrying Conductor in a Magnetic Field
Fleming's left hand rule
Domestic circuits
Domestic electric circuits

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Magnetic force on a current carrying wire: Tutorial and Examples

Magnetic force on current-carrying wire problems with full explanations are provided for high school physics.

Introduction

Experiments show that when a wire carrying a current is immersed in a uniform magnetic field, a deflecting force is applied to it. But what is the physical origin of this force?

Recall that, once a moving charge encounters a magnetic field, a side-way force exerts on it. A wire-carrying current is made up of a very large number of such moving charges.

Let's go and see what happens inside a wire. Current in a wire is due to the movement of the conduction electrons (by convention, their movement is in the opposite direction of the current).

Suppose these electrons move toward the right with a drift velocity $v_d$. From the magnetic field, a force $\vec{F}_B$ of magnitude $ev_d B$ exerted on each such conduction electron.

The right-hand rule tells us that the direction of the force must be upward. Thus, we can conclude an upward force is applied to the wire due to an upward force exerted on each conduction electron inside it.

The above consequence is in full agreement with the experiment mentioned earlier.

Now, with this physics in mind, we want to derive the equation of force on a current-carrying wire in a magnetic field as below.

Consider a length $L$ of the wire immersed in a magnetic field directed out of the plane of the page and conduction electrons travel to the right. Those electrons with drift velocity $v_d$ take time $t=\frac{L}{v_d}$ to traverse the length $L$.

Thus, in that time a net charge of $q=it=\frac{iL}{v_d}$ passes through the end of the length. The magnetic force on this moving charge is \begin{align*} F_B&=q v_d B \sin \theta \\ \\ &=\Big(\frac{iL}{v_d}\Big)v_d B \sin 90^\circ\\ \\ &= iLB\end{align*} in above, $\theta = 90^\circ$ substituted since electrons move at a right angle with the magnetic field.

Therefore, the force on a straight wire carrying current $i$ and immersed in a uniform magnetic field $B$ that is perpendicular to the wire is determined as $F_B=i L B$.

We can generalize the above formula, to include the case in which the magnetic field is not perpendicular to the wire as \[ \vec{F}_B = i\vec{L}\times \vec{B} \] In this cross product, $\vec{L}$ is the length vector with a magnitude of length of that segment of the wire inserted into the magnetic field. The direction of this vector is directed along the direction of the current.

The magnitude of the force is also given as $F_B= i LB \sin \theta$, where $\theta$ is the angle between $\vec{L}$ and $\vec{B}$.

The direction of the $\vec{F}_B$ is perpendicular to the plane of $\vec{L}$ and $\vec{B}$ and is determined using the right-hand rule. Put your right four fingers along the current direction, so that your palm is directed toward the magnetic field direction, and your thumb shows the direction of the magnetic force.

In the following, some problems are presented for a better understanding of this topic

Magnetic force on a current-carrying wire problems

Problem (1): A current $i$ passes through a wire and is immersed in a uniform magnetic field B so that a maximum force applies to it, as shown in the figure below. Find the direction of the magnetic field.

A current carrying wire in a magnetic field problem

Solution : Method (I) using the right-hand rule Put your right four fingers in the direction of the current $i$ so that the thumb points to the magnetic force. In this setup, your palm which is directed to the negative of the $z$-axis shows the direction of the magnetic field.

Method (II) using cross-product algebra The equation of the magnetic force on a current-carrying wire is $\vec{F}_B=i \vec{L}\times \vec{B}$. In this example, the length vector $\vec{L}$ is toward the negative of y-axis that is $\vec{L}=L\,(-\hat{j})$ and force is to the x-axis , $\vec{F}_B=F_B\,(\hat{i})$. Substituting these vectors in the above formula, we have \begin{align*} \vec{F}_B &= i\vec{L}\times \vec{B} \\ F_B\,(\hat{i}) &= L\,(-\hat{j}) \times \vec{B}\end{align*} Thus, $\vec{B}$ must be directed to $-\hat{k}$ or $\hat{k}$ (since maximum force is obtained when force is perpendicular to both $L$ and $B$). From the cross-product algebra, one can deduce $B$ must be to $-\hat{k}$ to produce a vector along the $x$-axis.

Problem (2): A wire is extended horizontally and a constant current passes through it toward the east. The Earth's magnetic field is from south to north. What is the direction of the force applied to it from the earth's magnetic field?

Solution : first realize that, in a plane, directions are as follows: east is to the right, west to the left, north into the page, south out of the page, and up and down is trivial (see figure below).

direction of a current-carrying wire problem in an external magnetic field

Thus, using the right-hand rule, the force on the wire is in the up direction.

Solution : put the given data into the equation of force on a current carrying wire $F_B= i L B \sin \theta$ as follows \begin{align*} F_B &= i L B \sin \theta \\ &= 25\times 0.8 \times (500\times 10^{-4}) \times \sin 37^\circ \\ &= 0.6\quad {\rm N}\end{align*} Note that the $SI$ unit of the magnetic field is Tesla and recall that ${\rm 1\, G= 10^{-4}\, T}$.

Use the right-hand rule to find the direction of the magnetic force on a positive charge moving in a uniform $B$ . Put your right fingers in the direction of the current and through the smaller angle curl them toward the magnetic field. The thumb points to the force direction. In this case, the force is into the page as $\otimes$.

Problem (4): A long and straight wire carrying the current i = 5 A is immersed in the magnetic field $\vec{B}=0.18\,\hat{i}+0.24\,\hat{j} \quad {\rm T}$. Find the maximum magnetic force on each meter of this wire.

Solution : Remember from the magnetic field problems section that the magnitude of the force exerted on a wire by a magnetic field is maximum when the direction of the wire is perpendicular to the field. Thus, the magnitude of the maximum force $F_{max} = i \ell B $ obtained as \begin{align*}F_{max} &= i \ell B \\ &=5\times 1\times 0.3 \\&= 1.5\quad {\rm N}\end{align*} where in above the magnitude of magnetic field vector is determined as below \begin{align*} B&=\sqrt{B_x^2+B_y^2}\\&=\sqrt{(0.18)^2+(0.24)^2}\\&=0.3\quad {\rm T}\end{align*}

Example (5): A straight wire of length $\ell=5\pi$ is formed into a semi-circle as shown in the figure below and placed in an external magnetic field of $B = 25\,{\rm G}$. If a current of $i=2\,{\rm A}$ passes through it, find the magnetic force on the wire due to the external magnetic field.

Solution : Recall that in the equation of magnetic force on a wire carrying current, $\vec{F}_B=i\vec{L} \times \vec{B}$, $\vec{L}$ was the length vector (or displacement vector ) which extends from the initial point to the final point on the wire.

Here, the length vector $\vec{L}$ extends from left to right in the direction of the x-axis i.e. $\vec{L}=L\,\hat{i}$ where $L$ is equal to the diameter of the semi-circle and its magnitude is calculated as \begin{align*} \text{length of wire} &= \text{circumference of semi-circle} \\ 5\pi &= \pi \, r\\ \Rightarrow r&= 5\,{\rm cm} \end{align*} Thus, the diameter is $L=2r=10\,{\rm cm}$.

Now, we can find the magnitude of the force as \begin{align*} F &= i L B \sin \theta \\ &=2\times (5\times 10^{-2})\times (25\times 10^{-4})\times \sin 90^\circ \\ &= 5\times 10^{-3}\quad {\rm N}\end{align*}

Direction of magnetic force on a wire inside B

Solution : the current flows out of the plane of the page, i.e. $\odot$, and the magnetic field is also directed from the west(left) to the east(right). Therefore, the force is exerted upward.

Right hand rule for direction of magnetic force on a wire

Problem (7): In the figure below, a piece of ABCDE wire is immersed into a uniform magnetic field B. What current must pass through the wire to exert a net force of 2 N from the magnetic field on the wire?

Solution : there are two methods to solve such problems. When a straight wire is formed into an arbitrary shape and inserted into a magnetic field, the easiest way to find the force on it is to draw the length vector $\vec{L}$ and then find the force on this vector.

Here, the length vector is plotted directly from A to E. This vector has two components, one is DE and the other is AD. The former is perpendicular to $B$ and the latter is parallel so its contribution to magnetic force is zero.

Thus, the DE part only contributes to the magnetic force since the angle between it and $B$ is $90^\circ$. Putting all these data into the equation $F_B=i\ell B \sin \theta$, we get the force on the whole ABCDE wire as below \begin{align*} F_B &=i\ell B \sin \theta \\ 2 &= i\,(0.1)\times 4\times \sin 90^\circ \\ \Rightarrow i&= 5\quad {\rm A}\end{align*}

Problem (8): in a part of space there is a uniform magnetic field $\vec{B}=12\,\hat{i}+5\,\hat{j}$. A long and straight wire carrying current $i=3\,{\rm A}$ is placed into that area at the right angle with the field. What is the net magnetic force on each meter of the wire?

Solution : first find the magnitude of the magnetic field as below \begin{align*} B&=\sqrt{B_x^2+B_y^2}\\&=\sqrt{12^2+5^2}\\ &=13 \quad {\rm T}\end{align*} next calculate the force on the wire using the equation of maximum magnetic force on a straight wire carrying current $i$ as \begin{align*} F_{max} &=i\ell B\\ &=3\times 13\times 1\\&=39\quad {\rm N}\end{align*} since the wire is perpendicular to the magnetic field so the force on it is maximum.

The magnitude of the force on a wire carrying current i in an arbitrary magnetic field B is F = i L B sin θ, where θ is the smaller angle between the magnetic field and current.
The direction of the force is determined by the right-hand rule: put your right fingers along the direction of the current such that your palm points to the magnetic field and your thumb shows the force's direction.
The maximum magnetic force occurs when the angle between B and i be θ = 90°.
The force on an arbitrary part of a bent wire is the same as the force on a straight wire carrying the same current and extends between the two endpoints of that section.

Author: Dr. Ali Nemati

Date Created: 1/18/2021

11.4 Magnetic Force on a Current-Carrying Conductor

Learning objectives.

By the end of this section, you will be able to:

Determine the direction in which a current-carrying wire experiences a force in an external magnetic field
Calculate the force on a current-carrying wire in an external magnetic field

Moving charges experience a force in a magnetic field. If these moving charges are in a wire—that is, if the wire is carrying a current—the wire should also experience a force. However, before we discuss the force exerted on a current by a magnetic field, we first examine the magnetic field generated by an electric current. We are studying two separate effects here that interact closely: A current-carrying wire generates a magnetic field and the magnetic field exerts a force on the current-carrying wire.

Magnetic Fields Produced by Electrical Currents

When discussing historical discoveries in magnetism, we mentioned Oersted’s finding that a wire carrying an electrical current caused a nearby compass to deflect. A connection was established that electrical currents produce magnetic fields. (This connection between electricity and magnetism is discussed in more detail in Sources of Magnetic Fields .)

The compass needle near the wire experiences a force that aligns the needle tangent to a circle around the wire. Therefore, a current-carrying wire produces circular loops of magnetic field. To determine the direction of the magnetic field generated from a wire, we use a second right-hand rule. In RHR-2, your thumb points in the direction of the current while your fingers wrap around the wire, pointing in the direction of the magnetic field produced ( Figure 11.11 ). If the magnetic field were coming at you or out of the page, we represent this with a dot. If the magnetic field were going into the page, we represent this with an × . × . These symbols come from considering a vector arrow: An arrow pointed toward you, from your perspective, would look like a dot or the tip of an arrow. An arrow pointed away from you, from your perspective, would look like a cross or an × . × . A composite sketch of the magnetic circles is shown in Figure 11.11 , where the field strength is shown to decrease as you get farther from the wire by loops that are farther separated.

Calculating the Magnetic Force

Electric current is an ordered movement of charge. A current-carrying wire in a magnetic field must therefore experience a force due to the field. To investigate this force, let’s consider the infinitesimal section of wire as shown in Figure 11.12 . The length and cross-sectional area of the section are dl and A , respectively, so its volume is V = A · d l . V = A · d l . The wire is formed from material that contains n charge carriers per unit volume, so the number of charge carriers in the section is n A · d l . n A · d l . If the charge carriers move with drift velocity v → d , v → d , the current I in the wire is (from Current and Resistance )

The magnetic force on any single charge carrier is e v → d × B → , e v → d × B → , so the total magnetic force d F → d F → on the n A · d l n A · d l charge carriers in the section of wire is

We can define dl to be a vector of length dl pointing along v → d , v → d , which allows us to rewrite this equation as

This is the magnetic force on the section of wire. Note that it is actually the net force exerted by the field on the charge carriers themselves. The direction of this force is given by RHR-1, where you point your fingers in the direction of the current and curl them toward the field. Your thumb then points in the direction of the force.

To determine the magnetic force F → F → on a wire of arbitrary length and shape, we must integrate Equation 11.12 over the entire wire. If the wire section happens to be straight and B is uniform, the equation differentials become absolute quantities, giving us

This is the force on a straight, current-carrying wire in a uniform magnetic field.

Example 11.4

Balancing the gravitational and magnetic forces on a current-carrying wire, significance, example 11.5, calculating magnetic force on a current-carrying wire.

We start with the general formula for the magnetic force on a wire. We are looking for the force per unit length, so we divide by the length to bring it to the left-hand side. We also set sin θ = 1 . sin θ = 1 . The solution therefore is F = I l B sin θ F l = ( 5.0 A ) ( 0.30 T ) F l = 1.5 N/m. F = I l B sin θ F l = ( 5.0 A ) ( 0.30 T ) F l = 1.5 N/m. Directionality: Point your fingers in the positive y -direction and curl your fingers in the positive x -direction. Your thumb will point in the − k → − k → direction. Therefore, with directionality, the solution is F → l = −1.5 k → N/m. F → l = −1.5 k → N/m.
The current times length and the magnetic field are written in unit vector notation. Then, we take the cross product to find the force: F → = I l → × B → = ( 5.0 A ) l j ^ × ( 0.30 T cos ( 30 ° ) i ^ + 0.30 T sin ( 30 ° ) j ^ ) F → / l = −1.30 k ^ N/m. F → = I l → × B → = ( 5.0 A ) l j ^ × ( 0.30 T cos ( 30 ° ) i ^ + 0.30 T sin ( 30 ° ) j ^ ) F → / l = −1.30 k ^ N/m.

Check Your Understanding 11.3

A straight, flexible length of copper wire is immersed in a magnetic field that is directed into the page. (a) If the wire’s current runs in the + x -direction, which way will the wire bend? (b) Which way will the wire bend if the current runs in the – x -direction?

Example 11.6

Force on a circular wire.

where θ θ is the angle between the magnetic field direction (+ y ) and the segment of wire. A differential segment is located at the same radius, so using an arc-length formula, we have:

In order to find the force on a segment, we integrate over the upper half of the circle, from 0 to π . π . This results in:

The lower half of the loop is integrated from π π to zero, giving us:

The net force is the sum of these forces, which is zero.

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Book URL: https://openstax.org/books/university-physics-volume-2/pages/1-introduction
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Force on a Current-Carrying Conductor ( CIE IGCSE Physics )

Revision note.

Physics Project Lead

Force on a current-carrying conductor

When interacting with an external magnetic field, it therefore will experience a force
A simple situation would be a copper rod placed within a uniform magnetic field
When current is passed through the copper rod, it experiences a force which makes it move

Copper rod experiment, downloadable IGCSE & GCSE Level Physics revision notes

A copper rod moves within a magnetic field when current is passed through it

reversing the direction of the current
reversing the direction of the magnetic field

This phenomenon is sometimes referred to as 'the motor effect'. The direction of the force is determined by Fleming's left-hand rule.

Flemings left-hand rule

the current
the magnetic field
the thumb points in the direction of the force, or thrust, on the conductor
the first finger points in the direction of the magnetic field
the second finger points in the direction of current flow (from positive to negative)

Fleming’s left-hand rule can be used to determine the directions of the force, magnetic field and current

This means that sometimes the force could be into and out of the page (in 3D)

Worked example

A current-carrying wire is placed into the magnetic field between the poles of the magnet, as shown in the diagram.

WE Flemings LHR Question Image, downloadable IGCSE & GCSE Physics revision notes

Use Fleming’s left-hand rule to show that there will be a downward force acting on the wire.

Step 1: Determine the direction of the magnetic field

Start by pointing your F irst F inger in the direction of the (magnetic) F ield

Step 2: Determine the direction of the current

Now rotate your hand around the first finger so that the se C ond finger points in the direction of the C urrent

Step 3: Determine the direction of the force

The TH umb will now be pointing in the direction of the TH rust (the force)
Therefore, this will be the direction in which the wire will move

WE Flemings LHR Answer Image, downloadable IGCSE & GCSE Physics revision notes

Remember that the magnetic field is always in the direction from North to South and current is always in the direction of a positive terminal to a negative terminal.

Feel free to use Fleming's left hand rule in your exam, just don't make it too distracting for other students!

Charged particles in a magnetic field

Extended tier only

This is because the magnetic field exerts a force on each individual electron flowing through the wire
The force is always at 90 degrees to both the direction of travel and the magnetic field lines
The direction can be worked out by using Fleming's left-hand rule
the second finger (current) points in the opposite direction to the direction of motion
this is because conventional current flows in the opposite direction to electron flow

Deflected particle, IGCSE & GCSE Physics revision notes

When a charged particle (such as an electron) enters a magnetic field, it is deflected by the field

it will experience the maximum force
it will experience no force
it will experience a small force

Remember that the direction of current is the direction of positive charged. Therefore, if a particle has a negative charge (such as an electron), then the second finger (current) must point in the opposite direction to its direction of travel.

The left-hand rule can be applied to any charged particles, but in the IGCSE exam questions are likely to stick to electrons.

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P14.5 – Force on a Current Carrying Conductor

Disclaimer: Due to unforeseen difficulties, we have had to take down the images on this notes page. They will be replaced shortly. We apologise for the inconvenience, but hope that the new images will provide you with an even better learning experience.

Describe and interpret an experiment to show that a force acts on a current-carrying conductor in a magnetic field, including the effect of reversing: • the current • the direction of the field

If a wire carrying a current is placed in a magnetic field (with lines of force at right angles to the wire) then it will experience a force at right angles to both the current direction and field lines.

The force on the wire is increased if:

The current is increased
A stronger magnet is used
The length of the wire in the field is increased
State and use the relative directions of force, field and current.

To predict the direction of the force, current or fields we can use Fleming’s left-hand rule:

So when reversing the current or direction of fields, the direction of force also changes. You can use the above rule to figure out the exact change. Try this question for practice and check the answer below.

(Hint: a magnetic field always runs from n orth to south)

ANSWER: Downwards Between the magnets, the current is flowing in the north direction, away from us (not the magnet’s north!), so you point your second finger away from you. The magnetic field runs from the north pole of the magnet to its south pole, i.e. from the left to the right and so point your forefinger in that direction. Don’t change the position of your second finger accidentally! And so naturally your thumb which indicates current will be in the downward position!

Notes submitted by Lintha

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Describe the activity that shows that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic field. How does Fleming’s left-hand rule help us to find the direction of the force acting on the current-carrying conductor?

Activity for a current-carrying conductor: take a little aluminium rod and write ab on it of about 5 centimeter. suspend it horizontally from a stand using two connecting cables. place a powerful horseshoe magnet between the two poles with the magnetic field pointing upwards. place the north pole of the magnet vertically below the aluminium rod and the south pole vertically above it for this. connect the aluminium rod to a battery, a key, and a rheostat in series. now, from end b to end a, run a current through the aluminium rod. it is observed that the rod is displaced towards the left. reverse the direction of the stream flowing through the rod to see which way it is displaced. it's moving right now. fleming left-hand rule: when an electric current flows through a straight wire and an external magnetic field is put across it, the wire experiences a force that is perpendicular to both the field and the direction of the current flow, according to fleming's left-hand rule. the direction of the force, magnetic field, and current may all be determined using this rule. the thumb, first finger, and second finger of the left hand can all be held perpendicular to each other. the direction of motion arising from the force on the conductor is represented by the thumb. the first finger represents the magnetic field's direction. the current's direction is represented by the second finger..

(a) A current-carrying conductor is placed perpendicularly in a magnetic field. Name the rule which can be used to find the direction of force acting on the conductor.

(b) State two ways to increase the force on a current-carrying conductor in a magnetic field.

(a) Draw a sketch to show the magnetic lines of force due to a current-carrying straight conductor.

(b) Name and state the rule to determine the direction of magnetic field around a straight current-carrying conductor.

Which rule helps to find the direction of force acting on a current-carrying conductor placed in a magnetic field. State the rule.

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(a) Explain an activity to show that a current-carrying conductor experiences a force when placed in a magnetic field. (b) State the rule which gives the direction of force acting on the conductor.. (c) An electron moves perpendicular to a magnetic field as shown in the figure. What would be the direction of force experienced on the electron?

(a) a small aluminium rod suspended horizontally from a stand using two connecting wires. place a strong horseshoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. for this, put the north pole of the magnet vertically below and south pole vertically above the aluminium rod. connect the aluminium rod in series with a battery, a key and a rheostat. pass a current through the aluminium rod from one end to other (b to a). the rod is displaced towards left. when the direction of current flowing through the rod is reversed, the displacement of rod across towards right. (b) fleming’s left- hand rule. stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular to one another. if the forefinger points in the direction of magnetic field and the middle finger in the direction of current, then the thumb will points in the direction of motion or the force acting on the conductor. (c) according to fleming's left hand rule, the direction of force is perpendicular to the direction of magnetic field and current. we know that the direction of current is taken opposite to the direction of motion of electrons. therefore, the force is directed upwards from the plane of the paper..