Magnetic flux is defined as the number of magnetic field lines crossing per unit area kept normal to the direction of lines of force.

When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

(i) Two equal and opposite charges separated by a small distance constitute an electric dipole.
(ii) The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them, \(|\vec{p}|=2 q a\).
(iii) The electric dipole moment vector lies along the line joining two charges and is directed from -q to +q.

Note: A similar problem is solved in XIth Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:
(i) The downward gravitational force exerted by the Earth (mg)
(ii) The normal force exerted by the inclined surface (N)
(iii) The Coulomb force given by uniform electric field (qE) The free body diagram for the mass m is drawn below.

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction.
Along x-direction, applying Newton’s second law, we have
mg sinθ\(\hat { i } \) - qE\(\hat { i } \) = 0
mg sinθ - q E = 0
or, E = \(\\ \frac { mgsin\theta }{ q } \)
Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.
The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.
E = \(\frac { mgh }{ qL } \).

V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5A\)

Microscopic form of ohm's law is 
\(\vec{J}=\sigma\vec {E}\) 
\(J=\frac{ne^2\tau}{m}\vec{E}\)
\(\frac{e \tau}{m} \rightarrow Drift \ velocity \ v_d\)
where \(\vec{J} \) - current density
\(\sigma \)  - conductivity
\(\vec{E} \) - Electric field

Gauss law:
(i) A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in Figure. then the total electric flux through the closed surface of the sphere is
\(\Phi_E =\oint { \vec { E } .d\vec { A } =\oint { Ed } Acos\theta } \)     .....(1) 

(ii) The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element \(d\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0^o.
\(\\ \Phi_E =\oint { EdA } \)  Since cos0^o = 1   ......(2)
iii) E is uniform on the surface of the sphere, 
\(\\ \Phi_E=E\oint { dA } \)  .......(3)
Substituting for \(\oint { dA=4{ \pi r }^{ 2 } } \) and \(E=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { r }^{ 2 } } \) in eqn (3), we get
\(\therefore \phi E=4{ \pi r }^{ 2 }E\)
\({ \phi }_{ E }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { Q }{ { r }^{ 2 } } \times { 4\pi r }^{ 2 }=4\pi \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } Q\)
\({ \phi }_{ E }=\frac { Q }{ { \varepsilon }_{ 0 } } \)   .....(4)
The equation (4) is called as Gauss's law.

Kirchhoff's First rule: (current rule)
(i) It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of law of conservation of electric charge.
(ii) All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. By convention current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying law to the junction A in Figure.

\({ I }_{ 1 }+{ I }_{ 2 }-{ I }_{ 3 }-{ I }_{ 4 }-{ I }_{ 5 }=0\)
(or)
\({ I }_{ 1 }+{ I }_{ 2 }=I_{ 3 }+{ I }_{ 4 }+{ I }_{ 5 }\)
Kirchhoff's Second rule (Voltage rule or Loop rule)
(i) It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit.
(ii) This rule follows from the law of conservation of energy for an isolated system (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

(iii) Kirchhof's voltage rule has to be applied only when all currents in the circuit reach a steady state condition.
(iv) The current in the various branches are constant. The product of current and resistance is taken as positive when the direction of the current is followed.
(v) Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Figure (a) and (b).
(vi) The emf is considered positive when proceeding from the negative to the positive terminal of the cell.

\(\frac { P }{ Q } =\frac { R }{ S } \), when the network is balanced
\(\frac { 500 }{ 800 } =\frac { x+400 }{ 1000 } \)
\(x+400=\frac { 5 }{ 8 } \times 1000\)
x + 400 = 625 
x = 625 – 400
x = 225 Ω

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
\(R=\frac { V^{ 2 } }{ P } =\frac { { (Ratedvoltage) }^{ 2 } }{ Ratedpower } \)
For 20W - 220V bulb,
\({ R }_{ 1 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =2420\Omega \)
For 100W - 220V bulb,
\({ R }_{ 2 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =484\Omega \)
Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, \(I=\frac { V }{ { R }_{ tot } } \)
R_tot (R₁ + R₂)
R_tot = (484 + 2420) \(\Omega\) = 2904 \(\Omega\)
\(I=\frac { 440V }{ 2904\Omega } \approx 0.151A\)
The voltage drop across the 20W bulb is
\(V_1=IR_1=\frac { 440V }{ 2904 }\times2420 \approx 366.6V\)
The voltage drop across the 100W bulb is
\({ V }_{ 2 }=I{ R }_{ 2 }=\frac { 440 }{ 2904 } 484\approx 73.3A\)
The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

Magnetic moment,
μ = IA
Angular momentum,
L = Iω
Ratio \(\frac{p_m}{L}=\frac{(q/T)\pi r^2}{mr^2\omega}=\frac{q}{2m}\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

Energy density u_E \(=\frac{U}{volume}\)
If A' = 2A     d ' = 2d
Then V ' = 2A  x 2d = 4Ad = 4V
Then volume would be increased. So, energy density will change.

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

(i) When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.
(ii) Let B be the magnetic field produced by passing current through the coil of the tangent galvanometer and B_H  be the horizontal component of earth's magnetic field.
(iii) Under the action of two magnetic fields, the needle comes to rest making angle with B_H , such that
B = B_H  tan \(\theta\)    .........(1)
Where B ⇒ magnetic field produced by current
B_H ⇒ horizontal component of earth's magnetic field
Construction:
(i) Copper coil of wire wound on a non-magnetic circular frame such as brass or wood. Compass box is kept at centre.
(ii) This compass box consists of pivoted magnet and aluminum pointer.
(iii) This compass box is having circular scale graduated with four quadrants.
Working:
(i) Two magnetic fields are perpendicular to each other.
(ii) Magnetic induction due to the current in the coil acting to normal to the plane of the coil.
(iii) Magnetic induction at the centre of the coil,
\(B=μ_o\frac{NI}{2R}\)   .....(2)
Sub. eqn.(1) in eqn. (2)
\(B_H tan \theta =μ_o\frac{NI}{2R}\)
\(B_H =μ_o\frac{NI}{2R}\frac{1}{tan\theta}\)

Unit of Inductance:
Inductance is a scalar and its unit is Wb A^-1 or Vs A^-1. It is also measured in henry (H).
1 H = 1 Wb A^-1 = 1 V sA^-1
The dimensional formula of inductance is [ML²T^-2A^-2].
If i = 1 A and NФ_B = 1 Wb turns, then L = 1 H.
Therefore, the inductance of the coil is said to be one henry, if a current of 1 A produces unit flux linkage in the coil.
If \(\frac { di }{ dt } =1\) As^-1 and ε = -1 V, then L = 1H
Therefore, One henry is defined as the inductance of the coil, if a current changing at the rate of 1 A s^-1 induces an opposing emf of 1 V in it.

(b) Assertion and Reason are true but Reason is the false explanation of the Assertion.

(a) Assertion and, Reason are correct and Reason is the correct explanation of Assertion.

(a) Assertion and Reason are correct and Reason is the correct explanation of Assertion.