Moving Charges and Magnetism: Force, Fields, Laws and Formula

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Moving charges and magnetism are two fundamental physics concepts that are related to each other.

  • Moving charges produces an electric field and the rate of flow of charge is known as current.
  • Magnetism is a physical property that occurs due to a magnetic field, allowing objects to attract or repel each other.
  • Magnetism due to electric current in a conductor is called Electromagnetism.
  • In 1820, Hans Christian Oersted demonstrated that both electricity and magnetism are related to each other.
  • The phenomena observed with needle alignment were the source of the relation between the moving charges and magnetism.
  • The needle's alignment was found to be tangent to an imagined circle with a straight wire in the center and a plane of the circle perpendicular to the wire.
  • When the current is applied, the orientation of the needle changes.
  • The passage of charges is supposed to produce the generation of a magnetic field.

Key Takeaways: Magnetism, Electric Field, Magnetic Field, Moving Coil Galvanometer, Solenoid, Ampere Circuital Law, Electric Current, Electricity, Wire, Electromagnetism


Magnetism

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The relationship between a Moving Charge and Magnetism is that Magnetism is produced by the movement of charges.

  • Magnetism is a property observed in Magnets and caused by moving charges that result in objects being attracted or pushed away.
  • The Oersted Experiment was a demonstration by Hans Christian Oersted that there was a connection between electricity and magnetism.
  • In this experiment, a current was switched on through a wire, it made a compass needle turn so that it was at right angles to the wire.
  • It means that an electric current creates a Magnetic Field.

moving charges and magnetism

Magnetism

Moving Charges and Magnetism:  Important Formulae

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Important formulas related to Moving charges and Magnetism iare tabulated below.

Parameter Formula
Lorentz Force

 \(F =q ( V \times B+E)\)

   Where,

V = velocity

E Electric Field.

B = Magnetic Field.

F = Force on a charge.

Ampere’s Circuital Law  \(\oint B. dl = \mu_0 I\)
Magnetic force on current-carrying conductor \(\vec F=I(\vec L \times \vec B)=BIL sin\theta\)
Angular frequency

\(\omega = 2 \pi f = \frac{qB}{ m}\)

     where

ω = angular frequency

= Frequency of rotation.

Motion of charged particle in a magnetic field \(r = \frac{mvsin \theta}{qB}\)
Oscillation Formula \(T = \frac{ 2 \pi}{\omega} = \frac{1}{f}\)

Magnetic Field

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The region or space around a magnet where the magnet has its magnetic effect is called the Magnetic field of the Magnet.

  • The strength of the magnetic field is also known as magnetic flux density, magnetic induction, or magnetic vector.
  • Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t) in the presence of both the electric field E (r) and the magnetic field B (r).
  • The force on an electric charge q due to both of them can be written as,

F = q [ E (r) + v x B (r)] = FElectric + Fmagnetic 

This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force.

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Magnetic Force

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The magnetic force is a result of the electromagnetic force, one of nature's four basic forces, and is produced by the motion of the charges.

  • When a charge moves, it creates a Magnetic field.
  • The force created in a Magnetic field is called Magnetic Force.
  • It is the basic force behind phenomena such as the action of electric motors and the attraction of magnets to iron.
  • Electric forces occur among electric charges at rest, while magnetic and electric forces exist between moving electric charges.
  • The magnetic force between two moving charges can be defined as the influence of one charge's magnetic field on the other.
  • Charge is the basic property associated with matter due to which it produces and experiences electrical and Magnetic effects.

The magnetism force experienced by a charge q moving in a uniform magnetic field B with velocity v is given by

\(F = q(\vec v \times \vec B)=qvBsin \theta\)

Where

  • F is the magnetic force
  • v is the velocity of the charge
  • q is the magnitude of the moving charge
  • B is the magnetic field
  • θ is the angle between v and B

Effect of Magnetic Field

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Various effects of the magnetic field on a current-carrying conductor are explained below.

Magnetic Force on a Current-Carrying Conductor

Consider a conductor that is carrying current.

  • Due to the motion of charges, every charge experiences a force. 
  • When a current is passed through a magnetic field, the magnetic field exerts a force on the wire in a direction perpendicular to both the current and the magnetic field.
  • The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule.

Let I be the current flowing through a conductor of length L placed in uniform magnetic field B, then the magnetic force experienced by the conductor is given by

\(\vec F=I(\vec L \times \vec B)=BIL sin\theta\)

  • If θ = 0 or 180, then F = 0. It means that the current-carrying conductor experiences no force when placed parallel or antiparallel to the direction of the magnetic field.
  • If θ = 90, then F = BIL. It means a current-carrying conductor experiences a maximum force when placed at the right angle to the uniform magnetic field.

Magnetic Force on a Current Carrying Conductor

Magnetic Force on a Current-Carrying Conductor

Motion in a Magnetic Field

In a Magnetic Field, the kinetic energy and speed of a charged particle (a particle that has current) is constant.

  • There is a magnetic force that acts perpendicular to the velocity of the particle.
  • This Magnetic Force is the reason to cause Circular Motion inside a Magnetic Field.

Consider a charge q moving with velocity v at a right angle in a uniform magnetic field B, the magnetic force experienced by the charge is given by

\(F=qvB\)

The radius of the circular path is given by

\(r=\frac{v}{(\frac {q}{m})B}\)

The frequency of the revolution is given by

\(f = \frac{qB}{2\pi m}\)

Motion in a Magnetic Field

Motion in a Magnetic Field

Read: Lorentz Force

Motion in Combined Electric and Magnetic Fields

The motion in a Combined Electric and Magnetic Fields depends on various factors:

  1. Velocity in Combined Electric and Magnetic Fields.
  2. Force
  3. Cross fields (when Electric and Magnetic Fields are perpendicular to each other)
  4. Perpendicular Velocity

Magnetic Field due to a Current Element

  • Magnetic field due to Current element is given by Biot-Savart’s Law.
  • According to this law, the Magnetic Field is proportional to the current, the element length, and inversely proportional to the square of the distance.
  • Direction is perpendicular to its objects.

According to Biot-Savart’s Law

\(dB=\frac{\mu_o}{4 \pi} \frac {Idl sin \theta}{r^2}\)

Where

  • dB is the strength of the magnetic field
  • I is the current flowing through element dl
  • r is the distance between the point where the magnetic field is to be calculated and the current-carrying element

Magnetic Field on the Axis of a Circular Current Loop

The magnetic on the axis of a current-carrying circular loop of radius R at distance x is given by

\(B=\frac{\mu_o}{4 \pi} \frac {2\pi IR^2}{(R^2+x^2)^{3/2}}\)

The magnetic field at the center of the current-carrying circular loop is given by

\(B=\frac {\mu_oI}{2R}\)


Ampere's Circuital Law

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Ampere’s Circuital Law states the relationship between an integrated magnetic field around a closed loop and the electric current passing through the loop.

  • It helps to count Magnetic Fields.
  • It uses high Symmetry during this process.

Ampere’s circuital law states that the line integral of the magnetic field around any closed path in free space is equal to the absolute permeability times the net current passing through any surface enclosed by the closed path.

Mathematically,

\(\oint \vec B.\vec{dl}=\mu_oI\)

Ampere’s Circuital Law Video Explanation


Soleniod and Toroid

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Solenoid: It is a helical structure. In a Solenoid, the Magnetic Field is Uniform. The field at the exterior near the solenoid is weak. Field lines in a solenoid are not seen near the vicinity. The inside field is parallel to its axis.

Toroid: It is a loop-like Structure similar to a hollow circular ring surrounded by turns of enameled wires. Magnetic Field is zero in Toroid and its direction is Clockwise.

Soleniod and Toroid

Solenoid and Toroid


Force on Current Carrying Conductors

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Let two straight conductors carrying currents I1 and I2 be placed at distance d. The magnetic force experienced by the conductors is given by

\(F=\frac{\mu_o}{4\pi}\frac{2I_1I_2}d{}\)

Force between Two Parallel Currents

There are two types of Force between two parallel currents:

  • Attractive: If the currents have the same direction or Currents are flowing in the same direction.
  • Repulsive: If the currents are in the opposite direction or they are flowing in the opposite direction.

          This force is responsible for the Pinch Effect. That’s why it's considered an important force.

Torque on Current Loop

  • When inside an electric motor, current carries a rectangular loop, the current passing through this loop produces a magnetic field which exerts a torque on the loop.
  • The whole process inside this loop produced a clockwise torque.
  • And this converts electrical energy into mechanical energy.

Torque on Current Loop

Torque on Current Loop


Moving Coil Galvanometer

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A Galvanometer is a device used to detect and measure small electric current in a circuit. It can measure current up to 10-6 A. The moving Coil Galvanometer is used to measure low electric currents. The moving Coil Galvanometer can be divided into:

  1. Suspended Coil Galvanometer: Current carrying coil, is acted upon by a torque because of its suspension in a uniform Magnetic Field. During the process of this torque, the coil moves in a circular round axis and the change in an object's velocity as a consequence of collision with the influence of a field (deflection) in a moving coil galvanometer is directly proportional to the current flowing through the coil.
  2. Pivoted- Coil or Weston Galvanometer: It is used in the laboratory to measure general or normal currents. It’s not used in sensitive cases.

Important Topics for JEE Main

As per JEE Main 2024 Session 1, important topics included in the chapter Moving Charges and Magnetism are as follows:

Topics Number of Questions Asked
Magnetic Field 5
Moving Coil Galvanometer 2
Effects of Magnetic Field 1

Things to Remember

  • Whenever we have a current carrying conductor, it produces a magnetic field
  • The intensity of a magnetic field can be measured mathematically using the Biot Savart law. It is the equivalent of coulomb’s law in electrostatics
  • Just like we have permittivity in electric charges, for magnetic fields we have the concept of permeability. 
  • Solenoid and Toroid are devices which apply the concept of electric field producing magnetic field. 
  • Since we can control the amount of current flowing through a conductor, electromagnets have greater applications than natural magnets. 

Also Read:


Previous Year Questions


Sample Questions

Ques. Delineate the trajectory of a charged particle moving with velocity v as it enters a uniform magnetic field perpendicular to the direction of its motion. (Comptt. All India 2012)

Ans.

moving charges and magnetism

The force acting on the charge particle will become perpendicular to both v and S for which it will describe a circular path. 

Ques. An ammeter of resistance 0.6 Ω can measure current upto 1.0 A. Find out (i) The shunt resistance that is required to enable the ammeter to measure current upto 5.0A.
(ii) The combined resistance of the shunt and the ammeter. (Delhi 2013)

Ans. (i) The shunt resistance, 

\(S = { R_AI_g \over i-i_g}\) = \({0.6 \times1 \over 4}\) = 0.15 \(\Omega\)

(ii) The total resistance,

\({1 \over R_{total}} = {1 \over 0.6} + {1 \over 0.15}\)

=0.12 \(\Omega\)

Ques. Define one ampere of current using the concept of force between two infinitely long parallel current carrying conductors. (All India 2013)

Ans.  \(F = {\mu_0\over 2\pi}{I_1I_2\over r}\)

One ampere of current refers to the value of steady current which when maintained in each of the two very long, parallel, straight conductors with negligible cross section and placed one meter above in vacuum. This would produce on each of the conductors a force which is equal to 2 x 10-7 newtons per metre (Nm-1) of length. 

Ques. Why the electrostatic field lines do not form closed loops? (All India 2015)

Ans. The electric field lines do not form a closed loop as the direction of an electric field is from positive to negative charge. Due to this one can consider a line of force starting from a positive charge and ending on a negative charge which indicates that electric field lines do not form closed loops.

Ques (a) How is toroid different from solenoid?
(b) Obtain the magnetic field inside a toroid by using Ampere’s critical law.
(c) Show that in an ideal toroid, the magnetic field is
(i) inside the toroid and 
(ii) outside the toroid at any point in the open space is zero. (Comptt. All India 2013)

Ans. (a) A toroid is a circular ring on which a wire is wound. While, a solenoid is a straight cylinder on which the wire is wound in a shape of helix. 

(b) A toroid is essentially a solenoid which is bent to form a ring shape. 

Let N be the number of turns per unit length of toroid and I be the current flowing in it.

Let N be the number of turns per unit length of toroid and I be the current flowing in it.

Assume a loop (region II) of radius r that passes through the centre of the toroid.

Let \(\overrightarrow{B} \) be the magnetic field along the loop,

 be the magnetic field along the loop,

Let (region II) B1 be the magnetic field outside toroid in open space. We can draw an amperian loop L2 of radius r2 through point Q.

Now, by applying ampere’s law,

\(\oint \overrightarrow{B_2 }dl = \mu_oI\) x No of Turns

B.2\(\pi\)r \(\mu_oIN\)

So, \(\overrightarrow{B} = {\mu_oI\over2\pi r}\)

As I = 0 and the circular turn current coming out of the plane of paper is cancelled exactly by current going into it, net I = 0.

As I = 0 and the circular turn current coming out of the plane of paper is cancelled exactly by current going into it, net I = 0.

(c) The Ampere’s critical law gives for the loop 1,

\(B_1.2\pi r_1 = \mu_0(0)\)

Therefore the magnetic field inside the toroid in open space is zero.

As I = 0 and the circular turn current coming out of the plane of paper is cancelled exactly by current going into it, net I = 0.

Also at point Q, we get

\(B_3(2\pi R_3) = \mu_0(I_{enclosed})\)

However from sectional cur, we refer to the current coming out of the plane of paper is cancelled exactly by the current going into it.

Thus, Ienclosed = 0

Therefore, B3 = 0

Ques. Draw a magnetic field line due to a current passing through a long solenoid. Use Ampere’s critical law to obtain the expression for the magnetic field due to the current I in a long solenoid with n number of turns per unit length. (Comptt. Delhi 2014)

Ans.

solenoid

Expression for magnetic field:

Let the length of solenoid be L

Total number of turns in solenoid be N

Number of turns per unit length is N/L = n

ABCD is an Ampere's loop where AB, DC are very large.

BC is in an area of 

\(\overrightarrow{B} = 0\)

AD is a long axis

Length of AD is x

Current in one turn is I0

Current in one turn is I0

By applying Ampere’s circuital loop, | B.d| = go I’

By applying Ampere’s circuital loop, | B.d| = go I’

Number of turns in x length is nx

Current in turn nx is I = nx I0

Based on Ampere’s critical law,

Bx = μ0 I ⇒ Bx = μ0nx I0

Therefore, B = μ0n I0

Ques. (a) Show how the Biot-Savart law can be alternatively expressed in the form of Ampere’s critical law. By using this law, obtain the expression for the magnetic field inside a solenoid of length l, cross sectional area A with N closely wound turns and carrying a steady current l. Draw the magnetic field lines of a finite solenoid carrying a steady current I.
(b) Straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at the ends. A current 5.0 A is set up in the rod through the wires. Find out the magnitude of the magnetic field which should be set up in. (Comptt. Delhi 2011)

Ans. Based on Biot-Savart law, we get to know that the magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point in which the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated.

 Based on Biot-Savart law, we get to know that the magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element

\(\left | dB \right |=(\frac{\mu _{0}}{4\pi })(\frac{Idlsin\Theta }{r^{2}})\)

The magnetic field on the axis of circular current loop,

The magnetic field on the axis of circular current loop,

The magnetic field on the axis of circular current loop,

In a special case, we may get the field at the centre of the loop where x = 0 and we obtain,

\({B_0 = \mu_0I\over{2R}}\)

In the current loop, both the opposite faces act as opposite poles making it a magnetic dipole. One side of the current carrying coil acts like the N-pole and the other side as the S-pole of a magnet. 

(b) Biot-Savart law can be alternatively expressed as the Ampere’s critical law by assuming the surface to be made up of a large number of loops. The sum of the tangential components of the magnetic field multiplied by the length of all the elements leads to integral. And according to Ampere’s critical law, this integral is equal to p0 times the total current passing through that surface,

\(\oint\overrightarrow{B.dl} = \mu_0I\)

(i) Expression for magnetic field within a solenoid-

Let number of turns per unit length be n

Total number of turns in the length h is nh

Thus, total enclosed current is nhl

Using Ampere’s circuital law,

Using Ampere’s circuital law,

(ii) Magnetic field lines of a finite solenoid,

(ii) Magnetic field lines of a finite solenoid,

(b) The magnetic field must be vertically inwards as per the figure shown to make tension zero. Therefore, force on the current carrying conductor and the weight of the conductor are equal and opposite which balance each other.

(b) The magnetic field must be vertically inwards as per the figure shown to make tension zero. Therefore, force on the current carrying conductor and the weight of the conductor are equal and opposite which balance each other.

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CBSE CLASS XII Related Questions

  • 1.
    The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.


      • 2.
        Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).


          • 3.
            The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

              • \( \frac{\alpha}{r} \)
              • \( \frac{\alpha r^2}{2} \)
              • \( \frac{\alpha}{2r^2} \)
              • \( -\frac{\alpha}{r} \)

            • 4.
              Answer the following giving reason:
              (a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
              (b) The saturation current in case (a) is different for different intensity.
              (c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.


                • 5.
                  A small spherical shell \( S_1 \) has point charges \( q_1 = -3 \, \mu C \), \( q_2 = -2 \, \mu C \) and \( q_3 = 9 \, \mu C \) inside it. This shell is enclosed by another big spherical shell \( S_2 \). A point charge \( Q \) is placed in between the two surfaces \( S_1 \) and \( S_2 \). If the electric flux through the surface \( S_2 \) is four times the flux through surface \( S_1 \), find charge \( Q \).


                    • 6.
                      Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
                      Three batteries E1, E2, and E3 of emfs and internal resistances

                        CBSE CLASS XII Previous Year Papers

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