Moving Charges and Magnetism Class 12 Notes CBSE Physics Chapter 4 (Free PDF Download)

Section-A (1 Marks Questions)

1. Is any work done on a moving charge by a magnetic field?

Ans. Since the magnetic force is perpendicular to the displacement of the moving charge, therefore the work done by the magnetic force is zero.

2. What is the gyromagnetic ratio?

Ans. Gyromagnetic ratio of an electron is the ratio of its magnetic moment to its orbital angular momentum.

$\mu _{l}=\dfrac{e}{2m_{c}}l$

μl is the magnetic moment of the electron, l is the oriental angular momentum of the electron.

$g(gyromagnetic\;ratio)=\dfrac{\mu _{l}}{l}=\dfrac{e}{2m_{c}}$

3. If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis, which way would the Magnetic force be for :

(a) an electron (negative charge), 

(b) a proton (positive charge)?  

Ans.

(a) For electron, the force will be along –z axis;

(b) For a positive charge (proton), the force is along the +z axis.

4. Why should the spring/suspension wire in a moving coil galvanometer have a low torsional constant?

Ans. Low torsional constant is basically required to increase the current/charge sensitivity in a moving coil ballistic galvanometer.

5. State one limitation of Ampere’s Circuital Law.

Ans. Ampere’s Circuital Law is only valid for steady currents. If the current changes with time then Ampere’s Circuital Law is not applicable.

Section – B (2 Marks Questions)

6. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer?

Ans. Two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer

(a) Non - Brittle conductor

(b) Restoring Torque per unit Twist should be small.

7. Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons?

Ans. We know that, Torque = NAIB

Since the area of circular loops is more than of a square loop torque experienced by a circular loop is greater.

8. A cyclotron is not suitable to accelerate electrons. Why?

Ans. A cyclotron is not suitable to accelerate electrons because its mass is less due to which they gain speed and step out of the dee immediately.

9. Write the two measures that can be taken to increase the sensitivity of a galvanometer.

Ans. Measures that can be taken to increase the sensitivity of a galvanometer,

• Increasing the no. of turns and the area of the coil,

• Increasing the magnetic induction and,

• Decreasing the couple per unit twist of the suspension wire.

10. Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.

Ans. The electric field and magnetic field should be perpendicular to each other. The velocity of the moving charge should be, v = E/B.

PDF Summary - Class 12 Physics Moving Charges and Magnetism Notes (Chapter 4)

1. Force on a moving charge:

The source of the magnetic field is a moving charge.

Suppose a positive charge $\text{q}$ is in motion in a uniform magnetic field $\overset{\to }{\mathop{\text{B}}}\,$ with velocity $\overset{\to }{\mathop{\text{v}}}\,$ .

$\text{n}$ 

$\therefore \,\,\text{F}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{qBvsin }\!\!\theta\!\!\text{ }\Rightarrow \text{F=kqBvsin }\!\!\theta\!\!\text{ }\,\,\left[ \text{k}=\text{constant} \right]$ 

Where in S.I. system, $\text{k}=1$ 

$\therefore \,\,\text{F=qBsin }\!\!\theta\!\!\text{ }\,\,\text{and}\,\,\overset{\to }{\mathop{\text{F}}}\,\text{=q}\left( \overset{\to }{\mathop{\text{v}}}\,\times \overset{\to }{\mathop{\text{B}}}\, \right)$ 

2. Magnetic field strength $\left( \overset{\to }{\mathop{\text{B}}}\, \right)$ :

We can see that in the equation,

$\text{F}=\text{qBvsin }\!\!\theta\!\!\text{ }$, if $\text{q}=1,\,\text{v}=1$,

$\text{sin }\!\!\theta\!\!\text{ }=1$ i.e. $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$ then $\text{F}=\text{B}$.

Therefore magnetic field strength can be known as the force felt by a unit charge in motion with unit velocity perpendicular to the direction of the magnetic field.

There are some special cases for this:

1. If $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$ or

${{180}^{\circ}}$, $\text{sin }\!\!\theta\!\!\text{ =0}$ 

$\therefore \,\,\text{F}=0$ 

A charged particle that is in motion parallel to the magnetic field will be not experiencing any force.

2. When $\text{v}=0,\text{F}=0$ 

At rest, a charged particle in a magnetic field will be not experiencing any force.

3. When $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }=1$ then the force will be maximum

${{\text{F}}_{\text{max}\text{.}}}=\text{qvB}$ 

A charged particle in a motion perpendicular to the magnetic field will be experiencing maximum force.

3. S.I. unit of magnetic field intensity: 

The S.I unit has been found to be tesla (T).

$\text{B}=\frac{\text{F}}{\text{qvsin }\!\!\theta\!\!\text{ }}$ 

When $\text{q}=1\text{C},\,\text{v}=\text{1m/s},\,\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$ That is, $\text{sin }\!\!\theta\!\!\text{ }=1$ and $\text{F}=\text{1N}$ 

Then $\text{B}=\text{1T}$.

At a point, the strength of the magnetic field can be called as \[\text{1T}\] if a charge of \[\text{1C}\] which have a velocity of $1$ m/s while in motion at the right angle to a magnetic field experiences a force of \[\text{1N}\] at that point.

4. Biot-Savart’s law:

The strength of magnetic flux density or magnetic field at a point \[\text{P}\](dB) because of the current element \[\text{dl}\] will be dependent on,

1. $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{I}$ 

2. $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{dl}$ 

3. $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{sin }\!\!\theta\!\!\text{ }$ 

4. $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\frac{\text{1}}{{{\text{r}}^{\text{2}}}}$,

When we combine them, $\text{dB}\,\text{ }\!\!\alpha\!\!\text{ }\,\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}\Rightarrow \text{dB=k}\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}$         [$\text{k}=$Proportionality constant]

In S.I. units, $\text{k}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}$ where ${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}$ can be called as permeability of free space.

${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}=\text{4 }\!\!\pi\!\!\text{ }\times \text{1}{{\text{0}}^{\text{-7}}}\,\text{T}{{\text{A}}^{\text{-1}}}\text{m}$ 

$\therefore \,\,\text{dB}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\frac{\text{Idlsin }\!\!\theta\!\!\text{ }}{{{\text{r}}^{\text{2}}}}$ and \[\overset{\to }{\mathop{\text{dB}}}\,=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\text{I}\frac{\left( \overset{\to }{\mathop{\text{dl}}}\,\times \overset{\to }{\mathop{\text{r}}}\, \right)}{{{\text{r}}^{\text{3}}}}\] 

$\overset{\to }{\mathop{\text{dB}}}\,$ will be perpendicular to the plane containing $\overset{\to }{\mathop{\text{dl}}}\,$ and $\overset{\to }{\mathop{\text{r}}}\,$ and will be directed inwards.

5. Applications of Biot-Savart’s law:

• Magnetic field $\left( \text{B} \right)$ kept at the Centre of a Current Carrying Circular Coil of radius $r$.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{2r}}$

If there are $n$ turns, then the magnetic field at the centre of a circular coil of n turns will be,

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}}{\text{2r}}$ 

Here \[\text{n}\] will be the number of turns of the coil. \[\text{I}\] will be the current in the coil and \[\text{r}\] will be the radius of the coil.

• Magnetic field because of a straight conductor carrying current.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{4 }\!\!\pi\!\!\text{ a}}\left( \sin {{\phi }_{2}}+\sin {{\phi }_{1}} \right)$ 

Here \[\text{a}\] will be the perpendicular distance of the conductor from the point where the field is to the measured.

${{\phi }_{1}}\,\,\text{and}\,\,{{\phi }_{2}}$ will be the angles created by the two ends of the conductor with the point.

In case of an infinitely long conductor, ${{\phi }_{1}}={{\phi }_{2}}=\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ 

$\therefore \,\,\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\text{.}\frac{\text{2I}}{\text{a}}$ 

• At a point on the axis, magnetic field of a Circular Coil Carrying Current.

If point \[\text{P}\] is lying far away from the centre of the coil.

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}\frac{\text{2M}}{{{\text{x}}^{\text{3}}}}$ 

Where $\text{M}=\text{nIA}=\text{magnetic}\,\text{dipole}\,\text{moment}\,\text{of}\,\text{the}\,\text{coil}$.

$\text{x}$ be the distance of the point where the field is needed to be measured, \[\text{n}\] be the number of turns, \[\text{I}\] be the current and \[\text{A}\] be the area of the coil.

• Magnetic field at the centre of a semi-circular current-carrying conductor will be,

$\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}}{\text{4a}}$

• Magnetic field at the centre of an arc of circular current-carrying conductor which is subtending an angle 0 at the centre will be,

$\text{B=}\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I }\!\!\theta\!\!\text{ }}{\text{4 }\!\!\pi\!\!\text{ a}}$

6. Ampere’s circuital law:

Around any closed path in vacuum line integral of magnetic field \[\overset{\to }{\mathop{\text{B}}}\,\] will be ${{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}$ times the total current through the closed path. that is, $\oint{\overset{\to }{\mathop{\text{B}}}\,\text{.}\overset{\to }{\mathop{\text{dl}}}\,}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{I}$ 

7. Application of Ampere’s circuital law:

1. Magnetic field because of a current carrying solenoid, $\text{B}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}$ 

$\text{n}$ be the number of turns per unit length of the solenoid.

In the edge portion of a short solenoid, $\text{B}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}}{\text{2}}$ 

2.  Magnetic field because of a toroid or endless solenoid

$\text{B}={{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}\text{nI}$ 

8. Motion in uniform electric field of a charged particle:–

Parabola is the path of a charged particle in an electric field.

Equation of the parabola be ${{\text{x}}^{\text{2}}}=\frac{\text{2m}{{\text{v}}^{\text{2}}}}{\text{qE}}\text{y}$ 

Where $\text{x}$ be the width of the electric field.

$\text{y}$ be the displacement of the particle from its straight path.

$\text{v}$ be the speed of the charged particle.

$\text{q}$ be the charge of the particle

$\text{E}$ be the electric field intensity.

$\text{m}$ be the mass of the particle.

9. In a magnetic field $(\overset{\to }{\mathop{B}}\,)$ which is uniform, the path of a particle which is charged in motion with a velocity $\overset{\to }{\mathop{\text{v}}}\,$ creating an angle $\text{ }\!\!\theta\!\!\text{ }$ with $\overset{\to }{\mathop{\text{B}}}\,$ will be a helix.

The component of velocity \[\text{vcos }\!\!\theta\!\!\text{ }\] will not be given a force to the charged particle, hence under this velocity in the direction of \[\overset{\to }{\mathop{\text{B}}}\,\], the particle will move forward with a fixed velocity. The other component $\text{vsin }\!\!\theta\!\!\text{ }$ will create the force $\text{F}=\text{qBvsin }\!\!\theta\!\!\text{ }$, which will be supplying the needed centripetal force to the charged particle in the motion along a circular path having radius $\text{r}$.

$\because \,\,\text{Centripetal}\,\text{force}=\frac{\text{m}{{\left( \text{vsin }\!\!\theta\!\!\text{ } \right)}^{\text{2}}}}{\text{r}}=\text{Bqvsin }\!\!\theta\!\!\text{ }$ 

$\therefore \,\,\text{vsin }\!\!\theta\!\!\text{ }=\frac{\text{Bqr}}{\text{m}}$ 

Angular velocity of rotation$=\text{w=}\frac{\text{vsin }\!\!\theta\!\!\text{ }}{\text{r}}=\frac{\text{Bq}}{\text{m}}$ 

Frequency of rotation\[=\text{v}=\frac{\text{ }\!\!\omega\!\!\text{ }}{\text{2 }\!\!\pi\!\!\text{ }}=\frac{\text{Bq}}{\text{2 }\!\!\pi\!\!\text{ m}}\] 

Time period of revolution$=\text{T=}\frac{\text{1}}{\text{v}}=\frac{\text{2 }\!\!\pi\!\!\text{ m}}{\text{Bq}}$ 

10. Cyclotron: 

This can be defined as a device we use for accelerating and therefore energize the positively charged particle. This can be created by keeping the particle, in an oscillating perpendicular magnetic field and a electric field. The particle will be moving in a circular path.

$\therefore \,\,\,\text{Centripetal}\,\text{force}=\text{magnetic}\,\text{Lorentz}\,\text{force}$ 

$\Rightarrow \frac{\text{m}{{\text{v}}^{\text{2}}}}{\text{r}}=\text{Bqv}\Rightarrow \frac{\text{mv}}{\text{Bq}}=\text{r}$ $\leftarrow $ radius of the circular path

Time for travelling a semicircular path$=\frac{\text{ }\!\!\pi\!\!\text{ r}}{\text{v}}=\frac{\text{ }\!\!\pi\!\!\text{ m}}{\text{Bq}}=\text{constant}$.

When ${{\text{v}}_{\text{0}}}$ be the maximum velocity of the particle and ${{\text{r}}_{\text{0}}}$ be the maximum radius of its path then we can say that,

$\frac{\text{mv}_{\text{0}}^{\text{2}}}{{{\text{r}}_{\text{0}}}}=\text{Bq}{{\text{v}}_{\text{0}}}\Rightarrow {{\text{v}}_{\text{0}}}=\frac{\text{Bq}{{\text{r}}_{\text{0}}}}{\text{m}}$ 

Maximum kinetic energy of the particle$=\frac{\text{1}}{\text{2}}\text{mv}_{\text{0}}^{\text{2}}=\frac{\text{1}}{\text{2}}\text{m}{{\left( \frac{\text{Bq}{{\text{r}}_{\text{0}}}}{\text{m}} \right)}^{\text{2}}}\Rightarrow {{\left( \text{K}\text{.E}\text{.} \right)}_{\text{max}\text{.}}}=\frac{{{\text{B}}^{\text{2}}}{{\text{q}}^{\text{2}}}\text{r}_{\text{0}}^{\text{2}}}{\text{2m}}$ 

Time period of the oscillating electric field$\Rightarrow \text{T}=\frac{\text{2 }\!\!\pi\!\!\text{ m}}{\text{Bq}}$.

Time period be the independent of the speed and radius.

Cyclotron frequency $=\text{v}=\frac{\text{1}}{\text{T}}=\frac{\text{Bq}}{\text{2 }\!\!\pi\!\!\text{ m}}$ 

Cyclotron angular frequency$={{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}=\text{2 }\!\!\pi\!\!\text{ v}=\frac{\text{Bq}}{\text{m}}$

11. Force acting on a current carrying conductor kept in a magnetic field will be,

\[\overset{\to }{\mathop{\text{F}}}\,=\text{I}\left| \overset{\to }{\mathop{\text{l}}}\,\times \overset{\to }{\mathop{\text{B}}}\, \right|\] or $\text{F}=\text{IlBsin }\!\!\theta\!\!\text{ }$ 

Here $\text{I}$ be the current through the conductor

$\text{B}$ be the magnetic field intensity.

$\text{l}$ be the length of the conductor.

$\text{ }\!\!\theta\!\!\text{ }$ be the angle between the direction of current and magnetic field.

1. If $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$ or ${{180}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }\Rightarrow 0\Rightarrow \text{F}=0$ 

$\therefore $  If a conductor is kept along the magnetic field, no force will be acting on the conductor.

2. If $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{sin }\!\!\theta\!\!\text{ }=1$, $\text{F}$ will be maximum.

${{\text{F}}_{\text{max}}}\text{=IlB}$ 

If the conductor has been kept normal to the magnetic field, it will be experiencing maximum force.

12. The force between two parallel current-carrying conductors:–

1.  If the current will be in a similar direction the two conductors will be attracting each other with a force

$\text{F}=\frac{{{\text{ }\!\!\mu\!\!\text{ }}_{\text{0}}}}{\text{4 }\!\!\pi\!\!\text{ }}.\frac{\text{2}{{\text{I}}_{\text{1}}}{{\text{I}}_{\text{2}}}}{\text{r}}$ per unit length of the conductor

2.  If the current is in opposite direction the two conductors will be repelling each other with an equal force.

3.  S.I. unit of current is \[1\] ampere. \[\left( \text{A} \right)\text{.}\] 

$\text{1A}$ can be defined as the current which on flowing through each of the two parallel uniform linear conductor kept in free space at a distance of $\text{1m}$ from each other creates a force of $\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}\,\text{N/m}$ along their lengths.

13. Torque experienced on a current carrying coil kept in a magnetic field:

$\overset{\to }{\mathop{\tau }}\,=\overset{\to }{\mathop{\text{M}}}\,\times \overset{\to }{\mathop{\text{B}}}\,\Rightarrow \tau =\text{MBsin }\!\!\alpha\!\!\text{ }=\text{nIBAsin }\!\!\alpha\!\!\text{ }$ where $\text{M}$ be the magnetic dipole moment of the coil.

 $\text{M}=\text{nIA}$ 

 Where $\text{n}$ be the number of turns of the coil.

 $\text{I}$ be the current through the coil.

 $\text{B}$ be the intensity of the magnetic field.

 \[\text{A}\] be the area of the coil.

\[\text{ }\!\!\alpha\!\!\text{ }\] will be the angle in between the magnetic field $\left( \overset{-}{\mathop{\text{B}}}\, \right)$ and normal to the plane of the coil.

Special Cases will be:

1. When the coil has been kept parallel to magnetic field $\text{ }\!\!\theta\!\!\text{ }={{0}^{\circ }}$, $\text{cos }\!\!\theta\!\!\text{ }=1$ then torque will be maximum.

${{\text{ }\!\!\tau\!\!\text{ }}_{\text{max}\text{.}}}=\text{nIBA}$ 

2. When the coil is kept perpendicular to magnetic field, $\text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$, $\text{cos }\!\!\theta\!\!\text{ }=0$ 

$\therefore \,\,\text{ }\!\!\tau\!\!\text{ }=0$ 

14. Moving coil galvanometer:

This has been on the basis on the principle that if a coil carrying current has been kept in a magnetic field it is experiencing a torque. There is a restoring torque because of the phosphor bronze strip which is bringing back the coil to its normal position.

In equilibrium,

\[\text{Deflecting torque }=\text{ Restoring torque}\] 

$\text{nIBA}=\text{k }\!\!\theta\!\!\text{ }$ [$\text{k}=\text{restoring torque/unit twist of the phosphor bronze strip}]\text{ }$ 

$\text{I}=\frac{\text{k}}{\text{nBA}}\text{ }\!\!\theta\!\!\text{ }=\text{G }\!\!\theta\!\!\text{ }$  where $\text{G}=\frac{\text{k}}{\text{nBA}}=\text{Galvanometer}\,\,\text{constant}$ 

$\therefore \,\,\text{I}\,\text{ }\!\!\alpha\!\!\text{ }\,\text{ }\!\!\theta\!\!\text{ }$ 

Current sensitivity of the galvanometer can be defined as the deflection made if the unit current has been passed through the galvanometer.

${{\text{I}}_{\text{s}}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{I}}=\frac{\text{nBA}}{\text{k}}$ 

Voltage sensitivity can be explained as the deflection created if unit potential difference has been applied across the galvanometer.

${{\text{V}}_{\text{s}}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{V}}=\frac{\text{ }\!\!\theta\!\!\text{ }}{\text{IR}}=\frac{\text{nBA}}{\text{kR}}$ $\left[ \text{R = Resistance of the galvanometer} \right]$ 

15. The maximum sensitivity of the galvanometer is having some conditions:- 

The galvanometer has been defined to be sensitive if a small current develops a large deflection.

$\because \,\,\text{ }\!\!\theta\!\!\text{ }=\frac{\text{nBA}}{\text{k}}\text{I}$ 

$\because \,\,\text{ }\!\!\theta\!\!\text{ }$ will be large if (i) $\text{n}$ is large, (ii) $\text{B}$ is large (iii) $\text{A}$ is large and (iv) $\text{k}$ is small.

16. Conversion of galvanometer into voltmeter and ammeter:

1. A galvanometer has been converted to voltmeter by putting a high resistance in series with it.

\[\text{Total resistance of voltmeter }=\text{ }{{\text{R}}_{\text{g}}}\text{ + R}\] where ${{\text{R}}_{\text{g}}}$ be the galvonometer resistance.

$\text{R}$ be the resistance added in series.

Current through the galvanometer$={{\text{I}}_{\text{g}}}=\frac{\text{V}}{{{\text{R}}_{\text{g}}}\text{+R}}$ 

Here $\text{V}$ is the potential difference across the voltmeter.

$\therefore \,\,\text{R}=\frac{\text{V}}{{{\text{I}}_{\text{g}}}}-\text{G}$ 

Range of the voltmeter: \[0\text{V volt}\text{. }\] 

2. A galvanometer can be converted into an ammeter by the connection of a low resistance in parallel with it (shunt)

$\text{Shunt}=\text{S}=\left( \frac{{{\text{I}}_{\text{g}}}}{\text{I}-{{\text{I}}_{\text{g}}}} \right){{\text{R}}_{\text{g}}}$ where ${{\text{R}}_{\text{g}}}$ be the galvanometer’s resistance.

$\text{I}$ be the total current through the ammeter.

${{\text{I}}_{\text{g}}}$  be the current through the ammeter. 

Effective resistance of the ammeter will be,

$\text{R}=\frac{{{\text{R}}_{\text{g}}}}{{{\text{R}}_{\text{g}}}+\text{S}}$ 

The range of the ammeter will be $0-\text{IA}$. An ideal ammeter will be having zero resistance.

Class 12th Physics Chapter 4 Notes Free PDF Download

Physics Class 12 Chapter 4 Notes illustrate the exertion of the magnetic field on the forces surrounding the current-carrying wires. It also shows how magnetic fields are produced with the help of currents. It is seen that within a cyclotron, there can be an acceleration of particles at relatively high energies.

The notes of Physics Class 12 Chapter 4 includes the sources, magnetic fields, motion within a magnetic field or in a combination of electric and magnetic fields. Apart from the laws that are discussed, there are ample derivations to take note of as well as mathematical sums on the topic. Solving this needs a clear understanding of the principles in this chapter. Students may download a free PDF of Class 12 Physics Chapter 4 Notes to prepare the topic better.

Moving Charges and Magnetism

The key concepts that are covered in the notes of Class 12 Physics Chapter 4 are:

1. Force on a Moving Charge

(image will be uploaded soon)

Magnetic field imposes a force on moving charges. The direction of the force that is exerted on a moving charge remains perpendicular to the plane that is formed which is denoted by 'v' and 'B' respectively. The 'Right-Hand Rule' is followed in this case. Moreover, the force's magnitude is in direct proportion to the sine of the angle between 'B' and 'v'.

2. Strength of Magnetic Field

Magnetic field strength is also known as magnetic field intensity. It includes such a component of a magnetic field which is a result of external current. It is denoted as a vector, H and the unit for measurement is amperes per metre. The equation is,

H = B/μ − M

[where,

B = magnetic flux density 

μ = magnetic permeability 

M = magnetisation] 

3. Biot Savart's Law

The law relates to current sources and magnetic fields. The determination of the magnetic field arises from the distribution of current involving vector products. It includes a problem in calculus where the distance among field points and current is constantly changing. Its major application includes calculating the magnetic impact on a molecular and atomic level and determining velocity in aerodynamics.

4. Ampere's Circuital Law

Ampere's Circuital Law encompasses the relationship between the magnetic field and its source current. The magnetic field density on an imaginary enclosed path amounts to be same as the product of current in the medium and permeability of the latter.

Moving charges and Magnetism Class 12 Notes also discuss cyclotron. It is engaged in the acceleration of charged atomic particles within a uniform magnetic field. It results in the production of radioactive isotopes which are utilised for imaging procedures.

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