PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

  Chapter–4: Moving Charges and Magnetism: Concept of magnetic field,  Oersted's experiment to show magnetic effects of electric current & Ampere's swimming Rule Magnetic Field and Magnetic Force State and Explain Lorentz Force  Force on a current-carrying conductor in a uniform magnetic field. Motion in a magnetic field  Magnetic field due to current element : Biot - Savart law  Magnetic field on the Axis of a circular current loop.  Ampere's law - Proof  its applications to infinitely long straight wire.  Force between two parallel current-carrying conductors. Force between two parallel current-carrying conductors. definition of ampere,  torque experienced by a current loop in uniform magnetic field. Current loop as a magnetic dipole and its magnetic dipole moment,  Moving coil galvanometer its principle , theory and construction. current and voltage sensitivity of galvanometer.  What is ammeter ? conversion Galvanom...

Derive an expression for magnetic force acting on a current carrying conductor placed in uniform magnetic field. A current carrying conductor contains large number of free electrons.These electrons move with drift velocity $\vec{v}_d$ in a direction opposite to the direction of conventional current flowing in the conductor. An electron moving in a uniform magnetic field $\vec{B}$ experiences a force which is transmitted to the conductor. Consider a conductor of length $l$ carrying a current $I$ placed in a uniform magnetic field $\vec{B}$. Let $n$ = Number of free electrons per unit volume of the conductor. Let $A$ = Area of cross-section of the conductor. Magnetic force acting on an electron is given by: $\vec{f}_m = -e (\vec{v}_d \times \vec{B}) \quad \dots (i)$ Now consider a small element of length $d\vec{l}$ of the given conductor. Number of electrons in the small element = n × volume of the element = $n (A dl)$. Magnetic force experienced by the element of the conductor is given ...

Charged particle moving in a uniform magnetic field  Case I. When motion of charged particle is parallel or antiparallel to the uniform magnetic field. In this case, the velocity $\vec{v}$ of the charged particle makes an angle $0^\circ$ or $180^\circ$ with the direction of the magnetic field. Therefore, magnetic force acting on the charged particle is given by $\vec{F}_m = q (\vec{v} \times \vec{B}) = qvB \sin \theta$ Since $\theta = 0^\circ$ or $180^\circ$,  so $\sin 0^\circ = 0$ or $\sin 180^\circ = 0$ $\vec{F}_m = 0$ Thus, moving charged particle does not experience any force. Therefore, the charged particle continues to move along a straight line if it moves parallel or anti-parallel to the direction of the uniform magnetic field. Case II. When charged particle moves at right angle to the magnetic field. Consider a uniform magnetic field ($\vec{B}$) acting perpendicular to the plane of the paper and directed into the paper. Let a charged particle having charge $+q$ enter ...

Magnetic Field on the Axis of a circular loop ( coil ) carrying Current  Consider a point P on the axis of the current loop at a distance $x$ from the centre O of the loop. Every current element ($d\vec{l}$) of the loop is perpendicular to the direction of $\vec{r}$ (represented by unit vector $\hat{r}$). Thus, $|d\vec{l} \times \hat{r}| = dl \sin 90^\circ = dl$.  Thus, every element of the current loop is at a distance $r = \sqrt{R^2 + x^2}$ from the point P on the axis of the current loop. According to Biot-Savart's law, the magnitude of the magnetic field due to current element ($Id\vec{l}$) at a distance $r$ is given by $ dB = \frac{\mu_0}{4\pi} \frac{I |d\vec{l} \times \hat{r}|}{r^2}$ $dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} $.... (1) The direction of the magnetic field is perpendicular to the plane formed by $d\vec{l}$ and $\hat{r}$ (along PL). Resolving magnetic field ($d\vec{B}$) into two components: (i) $dB_x = dB \cos \theta$ along the axis of the current loop (or c...

Magnetic Field due to Current Carrying Circular Wire of Infinite Length Using Ampere's law Consider a circular wire of infinite length and radius $a$ through which a steady current $I$ passes. The current in the wire gives rise to a magnetic field around it. The magnetic field lines are represented by concentric circles with their centres on the axis of the wire. magnetic field intensity due to the current carrying circular wire of infinite length at the points (i) out side the wire, (ii) on the surface of the wire and (iii) inside the wire. Case I. Magnetic field intensity at a point outside the wire. Consider a small portion of finite length of the wire of infinite length. Let $P_1$ be the point outside the wire at a distance $r$ ($> a$) from the axis of the wire. According to Ampere's circuital law, $ \oint \vec{B} \cdot d\vec{l} = \mu_0 I $ OR $ \oint B dl \cos 0^\circ = \mu_0 I \quad \text{or} \quad B \oint dl = \mu_0 I $ $ B \times 2 \pi r = \mu_0 I $ $ B = \frac{\mu_0...

What are the similarities between electric dipole and current loop ? Electric field intensity at a point on the axial line of a small electric dipole of electric dipole moment $p$ is given by,     $E = \frac{1}{4 \pi \varepsilon_0} \frac{2p}{x^3}$ where, $x$ is the distance of the point from the centre of the electric dipole. Magnetic field intensity at a point on the axis of a loop carrying current is given by     $B = \frac{\mu_0}{4 \pi} \frac{2 \pi I R^2}{(R^2 + x^2)^{3/2}}$ where, $R$ is the radius of the loop and $x$ is the distance of the point from the centre of the loop. But $\pi R^2 = A$ (Area of loop)     $B = \frac{\mu_0}{4 \pi} \frac{2 I A}{(R^2 + x^2)^{3/2}} \quad \dots (1)$ If $x >> R$, we have     $B = \frac{\mu_0}{4 \pi} \frac{2 I A}{x^3}$ Electric field intensity at a point on the equatorial line of small electric dipole is given by     $E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{x^3} \quad \dots (\text{iii})$ Magn...

Define Magnetic Dipole Moment of a Current Carrying Loop | SI unit and Dimensional formula . When current flows in a circular loop, magnetic field is set up around the current loop. The circular current loop behaves as a magnetic dipole. One face of the loop behaves as a North pole and the other face behaves as a South pole. The polarity of the face of circular current loop is determined by Clock Rule. If current round the face of loop or coil is in anticlockwise direction, then this face of the loop behaves as a North pole. If the current round the face of the loop is clockwise, then the face of the loop behaves as a South pole. Magnetic dipole moment ($\overrightarrow{m}$) of the current loop:  Magnetic dipole moment of a current loop is defined as the product of the current in the loop and the area of the loop. That is, m = IA If $r$ is the radius of the circular current loop then area of the loop, $A = \pi r^2$. Hence, magnetic dipole moment of the circular loop having current ...

What do you know about sensitivity of a galvanometer? Define current sensitivity and voltage sensitivity of a galvanometer. Sensitivity of a galvanometer :  A galvanometer is said to be sensitive if a small current flowing through the coil of galvanometer produces a large deflection in the galvanometer. (i) Current Sensitivity : The current sensitivity of a galvanometer is defined as the deflection produced in the coil of the galvanometer per unit current flowing through it. That is, current sensitivity  $= \dfrac{\phi}{I} = \dfrac{\phi}{k \phi / (NAB)} = \dfrac{NAB}{k}$ Current sensitivity of galvanometer can be increased either by (a) increasing the magnetic field B by using a strong permanent horseshoe shaped magnet. (b) increasing the number of turns N of the coil. But, number of turns of the coil cannot be increased beyond a certain limit. This is because the resistance of the galvanometer will increase subsequently and hence the galvanometer becomes less sensitive. (c) i...

State and Explain Lorentz Force ? - Electric force and Magnetic force  Lorentz Force : when a charged particle having charge q moves in a region , where both electric field $\overrightarrow{E}$ intensity and magnetic field $\overrightarrow{B}$ exist , it experiences a net force called Lorentz Force $\overrightarrow{F}$. Lorentz Force ,$\overrightarrow{F}$ = Force on charge due to electric field + Force on charge due to magnetic field  $$\overrightarrow{F}= \overrightarrow{F_{e}} +\overrightarrow{F_{m}}$$ (i) Electric Force ( $\overrightarrow{F_{e}}$) , A charged particle having charge q placed in an electric field $\overrightarrow{E}$ experience a force given by ,  Electric force - $$\overrightarrow{F_{e}} = q.\overrightarrow{E}$$ Direction of this force is same is same as that of electric field $\overrightarrow{E}$. (ii) Magnetic force ($\overrightarrow{F_{m}}$) :  Magnetic force Magnetic force on a charge q moving with velocity $\overrightarrow{v}$ at a certain ang...

Hans Christian Oersted performed a simple experiment to demonstrate the magnetic effects of electric current. (i) When current was allowed to pass through a wire AB placed along the axis of magnetic needle kept directly below and close to the wire that needle was found to deflect from its normal position. (ii) The deflection of the needle was found to be in the opposite direction on reversing the direction of the current by reversing the polarity of the battery.  (iii) Deflection of the magnetic needle change with the strength of the electric current. The deflection of the needle increase with the increase in current and vice versa  He concluded that an electric current (i.e. Flow of electric charges) in a conductor produce magnetic field in the space around the conductor. In another words flow of electric charges is the source of magnetic field. Ampere's swimming Rule :  Direction of the deflection of magnetic needle due to electric current in a conductor can be found by...

Magnetic field due to current carrying solenoid :  Consider a very long solenoid having n turns per unit length of solenoid.Let current I be flowing through the solenoid. The magnetic field inside the solenoid is almost uniform , strong and directed along. The axis of the solenoid. The magnetic field outside a very long solenoid is very weak and can be neglected.  Step 1 : Let P be a point well within the solenoid. Consider any rectangular loop ABCD ( known as Amperian Loop ) passing through P. Then  $\oint \overrightarrow{B}.\overrightarrow{dl}$ = Line integral of magnetic field across the loop ABCD,  $\oint \overrightarrow{B}.\overrightarrow{dl}= \int_{A}^{B} \overrightarrow{B}.\overrightarrow{dl} + \int_{B}^{C} \overrightarrow{B}.\overrightarrow{dl} + \int_{C}^{D} \overrightarrow{B}.\overrightarrow{dl} + \int_{D}^{A} \overrightarrow{B}.\overrightarrow{dl}$ $\overrightarrow{B}$ is perpendicular to paths BC and AD i.e. angle between $\overrightarrow{B}$ and $\overr...

Expression For Torque on current carrying rectangular loop placed in uniform magnetic field  Consider a rectangle conducting loop ABCD of a wire of length l and width b carrying current I ( clock wise ) is placed in a uniform magnetic field $\overrightarrow{B}$. The magnetic field $\overrightarrow{B}$ is acting X - axis along X-axis and the Normal of the plane of the loop $\hat{n}$ makes an angle $\theta$ with the magnetic field B.  The magnetic field B is perpendicular to the arms AB and CD of the loop and the angle between the arm BC or AD and the magnetic field is not 90°. Force acting on the arm AB of the conducting loop carrying current I due to magnetic field is given by :  $$\overrightarrow{F_{1}} = I(\overrightarrow{l}\times \overrightarrow{B})$$ According to Fleming's left hand rule , direction of $\overrightarrow{F_{1}}$ is perpendicular to the length of arm AB and directed into the plane of the paper  The magnitude of force $\overrightarrow{F_{1}}$ is give...

Sensitivity of a Galvanometer :  A galvanometer is said to be sensitive if a small current flowing through the coil of galvanometer produces a large deflection in the galvanometer.  (i) Current Sensitivity:  The current sensitivity of a Galvanometer is defined as the deflection produced in the coil of the galvanometer per unit current flowing through it. $$Current \ sensitivity = \frac{\phi}{I}$$ $$Current \ sensitivity = \frac{\phi}{\frac {k\phi}{NAB}}$$ $$Current \ sensitivity = \frac{\phi NAB}{k\phi}$$ $$Current \ sensitivity = \frac{NAB}{k}$$ Current sensitivity can be increased by :  (a) Increasing N  (b) Increasing B  (C) Increasing A (d) Decreasing k  (ii) Voltage sensitivity:  Voltage sensitivity of a Galvanometer is defined as the deflection produced in the coil of the galvanometer per unit voltage applied to it.  $$Voltage sensitivity= \frac{\phi}{V}$$ $$Voltage sensitivity= \frac{\phi}{IR}$$ $$Voltage\ sensitivity = \frac{\phi} {\...

Moving coil Galvanometer:  It is a device used to detect small current flowing in the electric circuit.  Principle :  Moving coil Galvanometer is based on the fact that when a current carrying loop or coil is placed in the uniform magnetic field , it experiences a torque. Construction:  it consists of a coil of copper wire wound on cylindrical soft iron core. The coil is pivoted in a uniform radial magnetic field provided by the concave shaped poles of a permanent magnet. The coil rotates freely about the pivot. A light pointer is attached to the coil. The pointer moves over a scale. A spring is attached to the coil to provide a restoring torque to the coil.  Theory :   Let B = Intensity of magnetic field  I = Current flowing through the coil L = length of coil b = Breadth of the coil N = Number of turns in the coil When current flows through the coil , it experiences a torque, which is given by : $$\tau = NIABsin \theta$$ Where , $\theta$ is the ...

Voltmeter : A voltmeter is an instrument used to measure the potential difference across the two ends of a circuit element.  Conversion of galvanometer into voltmeter :  A galvanometer can be converted into a voltmeter by connecting a large resistance in series to the galvanometer. Let G and R be the resistance of a Galvanometer and a conductor connected in series with it respectively. Let V volt be the potential difference to be measured by the voltmeter.Let $I_{g}$ be the current flowing in the circuit . Now potential difference between points A and B is given by : $$V = I_{g}R + I_{g}G$$ $$V = I_{g}(R+G)$$ $$\therefore \ R + G = \frac{V}{I_{g}}$$ $$R = \frac{V}{I_{g}}-G$$ Effective Resistance of the voltmeter :  Effective value of resistance of the galvanometer is given by : $$R^{'} = (R+G)$$ Which is very High Thus , voltmeter is a high resistance device , Resistance of an ideal voltmeter is infinite. Note : a voltmeter is always connected in parallel to the circuit c...

An ammeter is an instrument used to measure electric current in an electric circuit. A ammeter is a modified form of a galvanometer for measuring current in the circuit we require a device of a very low or practically zero resistance. Conversion of Galvanometer into Ammeter : "A Galvanometer can be converted into an ammeter by connecting a low resistance ( called shunt) parallel to the galvanometer."  A small resistance S (shunt resistance) connected in parallel to galvanometer (G) is shown in figure.  Let G and S be the resistance of a galvanometer and shunt respectively. Let I be the total current to be measured by an ammeter in the circuit.  Let $I_{g}$ be the current flowing through the galvanometer. The remaining current $(I-I_{g})$ flows through the shunt resistance.  Since G and S are parallel, the potential difference across them is same.  $$(I-I_{g}) S=I_{g} G$$ $$S = \left (\frac{I_{g} }{I-I_{g} } \right )G$$ This is the required value of shunt resist...

NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism 4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? Solution :   Given , n = 100 , r = 8 cm = 0.08m , I = 0.4 A  Magnitude of the magnetic field at the centre of the coil is given by the relation, $$\left | B \right | = \frac{\mu_{0} }{4\pi }.\frac{2\pi nI}{r}$$ where $\mu_{0} =4\pi \times10^{-7}TmA^{-1}$ $$B = \frac{4\pi \times 10^{-7}}{4\pi }.\frac{2\pi \times 100 \times 0.4}{0.08}$$ $$B = 10^{-7}. \frac{2\pi \times 10^{2} \times 4 \times 10^{-1} }{8 \times 10^{-2}}$$ $$B = \frac{2\pi \times 4}{8}.\frac{10^{-7} \times 10^{^{2}} \times 10^{-1}}{10^{-2}}$$ $$B = \pi \times 10^{-7+2-1+2}$$ $$B = 3.14 \times 10^{-4}T$$ 4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? Solution :  Given I = 35 A , r = 20 cm ...

Ampere's Circuital Law :   Ampere Circuital Law states that the line integral of the magnetic field around any closed path in free space is equal to absolute permeability ($\mu_{0}$) times the net current passing through any surface enclosed by the closed path.  Mathematically :  $$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}I$$ Where , $\overrightarrow{B}$ is the magnetic field , $\overrightarrow{dl}$ is the small element , $\mu_{0}$ is the absolute permeability of free space and I is the current enclosed by the closed path. Proof : Consider an infinitely long straight conductor carrying current I. The magnetic field lines are produced around the conductor as concentric circles. The magnetic field due to this current carrying infinite conductor at a distance a is given by $$B = \frac{\mu_{0}}{4\pi }(\frac{2I}{a})$$                            Consider a circle of radius a around the wire (cal...

Biot-Savart's law is used to determine the strength of the magnetic field at any point due to a current carrying conductor. Consider a very small element AB of length dl of a conductor carrying current I. The strength of magnetic field dB due to this small current element $(I\overrightarrow{dl})$ at point P, distant r from the element is found to depend upon the quantities as under: $$(i) dB \propto dl$$ $$(ii) dB \propto I$$ $$(iii) dB \propto sin\theta$$  where $\theta$ is the angle between $\overrightarrow{dl}$ and $\overrightarrow{r}$. $$(iv) dB \propto \frac{1}{r^{2}}$$ Combining (i) to (iv) , we get  $$dB \propto \frac{Idlsin\theta}{r^{2}}$$ $$dB = k \frac{Idl sin\theta}{r^{2}}$$ Where k is a constant of proportionality. In S.I units, $$k = \frac{\mu_{0}}{4\pi}$$ where , $\mu_{0}$ is called absoulte permeability of free space i.e . vacuum. Hence , equ (i) becomes  $$dB = \frac{\mu_{0}}{4\pi}. \frac{Idlsin\theta}{r^{2}}$$ Value of $\mu_{0}$ in S.I units = $4\pi \time...

Force between two infinitely long straight parallel conductors carrying currents -  ( i ) two infinitely long straight parallel conductors carrying currents in the same direction :  Consider two infinitely long parallel conductors X and Y carrying current $I_{1}$ and $I_{2}$ respectively in the same direction . let d be the distance of separation between these two conductor. The current $I_{1}$ in the conductor X produces a magnetic field around it. The magnitude of magnetic field at any point on the conductor Y due to current carrying in the conductor X is given by :  $$B_{1} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})$$ The direction of  $\overrightarrow{B}_{1}$  is perpendicular to the plane of the conductor Y and is directed into the plane as per right hand thumb rule , We know a current carrying conductor of length l placed as right angle to the magnetic field (B) experience a force is given by :  $$F =BIl$$ Therefore force experience per unit length of ...