Current flowing in wire A, I_A = 8.0 A
Current flowing in wire B, I_B = 5.0 A
Distance between the two wires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l = 10 cm = 0.1 m
Force exerted on length l due to the magnetic field is given as:
\(B=\frac{\mu_{0} 2 I_{A} I_{B} I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 8 \times 5 \times 0.1}{4 \pi \times 0.04}\)
= 2 x 10 ^-5 N
The magnitude of force is 2 x 10^–5 N. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

(a) Total resistance in the circuit is,
R_G+3 = 63 Ω. Hence, I = 3 / 63 = 0.048 A.
(b) Resistance of the galvanometer converted to an ammeter is,
\(\frac{R_{G} r_{s}}{R_{G}+r_{s}}=\frac{60 \Omega \times 0.02 \Omega}{(60+0.02) \Omega}=0.02 \Omega\)
Total resistance in the circuit is,
0.02Ω + 3Ω = 3.02Ω . Hence, I = 3 / 3.02 = 0.99 A.
(c) For the ideal ammeter with zero resistance,
I = 3 / 3 = 1.00 A

(a) dl and r for each element of the straight segments are parallel. Therefore, dl x r = 0. Straight segments do not contribute to |B|.
(b) For all segments of the semicircular arc, dl x r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 x 10^–4 T normal to the plane of the paper going into it.
(c) Same magnitude of B but opposite in direction to that in (b).

1u = 1.6605 x 10^–27 kg
To convert it into energy units, we multiply it by c² and find that energy equivalent = \(1.6605 \times 10^{-27} \times\left(2.9979 \times 10^{8}\right)^{2} \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}\)
\(=1.4924 \times 10^{-10} \mathrm{~J}\)
\(=\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
\(=0.9315 \times 10^{9} \mathrm{eV}\)
\(=931.5 \mathrm{MeV}\)
or, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
For, \({ }_{8}^{16} \mathrm{O}, \quad \Delta M=0.13691 \mathrm{u}=0.13691 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
\(=127.5 \mathrm{MeV} / \mathrm{c}^{2}\)
The energy needed to separate \({ }_{8}^{16} \mathrm{O}\) into its constituents is thus 127.5 MeV/c².

When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
\(R_{2}=\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}=85.8 \Omega\)
Using the relation
R₂ = R₁ [1 + α (T₂ - T₁)]
with α = 1.70 x 10^-4 °C^-1, we get
\(T_{2}-T_{1}=\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}=820^{\circ} \mathrm{C}\)
that is, T₂ = (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating effect due to the current equals heat loss to the surroundings) is 847 °C.

(a) \(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{7 \times(-2) \times 10^{-12}}{0.18}=-0.7 \mathrm{~J}\)
(b) W = U₂ – U₁ = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)=A \frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}+A \frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}=A \frac{7 \mu C}{0.09 m}+A \frac{-2 \mu C}{0.09 m}-0.7 \mathrm{~J}\)
= 70 − 20 − 0.7 = 49.3 J

(a) As \(V \propto \frac{1}{r}, V_{P}>V_{Q^{}}\) Thus, (V_P – V_Q) is positive. Also V_B is less negative than V_A . Thus, V_B > V_A or (V_B – V_A) is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)_A > (P.E.)_B and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

Initial current, I₁ = 5.0 A
Final current, I₂ = 0.0 A
Change in current dl = I^-1- I^-2 = 5A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
\(e=L\frac{di}{dt}\)
\(L=\frac{e}{\frac{di}{dt}}\)
\(=\frac{200}{\frac{5}{0.1}}=4H\)
Hence, the self induction of the coil is 4 H.

Number of turns in the solenoid, n = 800
Area of cross-section, A =  2.5 × 10^-4 m² 
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 x 3 x 2.5 x 10^-4
= 0.6 J T^-1

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square
 

 Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10 \(\sqrt{2}\) cm
AO = OC = DO = OB = 5  \(\sqrt{2}\) cm
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
\(tan2\theta =\frac { d }{ 1.5 } \)
∴ d = 1.5 x tan 70° = 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.

Refractive index of glass, μ
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R₁
Radius of curvature of the other face of the lens = R₂
Radius of curvature of the double-convex lens = R The value of R can be calculated as:
∴ R₁ = R and R₂ = -R
The value of R can be calculated as:
\(\frac { 1 }{ f } =(\mu -1)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
\(\frac { 1 }{ 20 } =(1.55)\left[ \frac { 1 }{ R } +\frac { 1 }{ R } \right] \)
\(\frac { 1 }{ 20 } =0.55\times \frac { 2 }{ R } \)
∴ R = 0.55 x 2 x 20 = 22 cm
Hence, the radius of curvature of the double-convex lens is 22 cm.

Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
(a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
\(r_{f b}=\Delta V / \Delta I=0.1 \mathrm{~V} / 10 \mathrm{~mA}=10 \ \Omega\)
(b) From the curve at V = –10 V, I = –1 \(\mu\) A,
Therefore,
\(r_{r b}=10 \mathrm{~V} / 1 \mu \mathrm{A}=1.0 \times 10^{7} \ \Omega\) 

Given, N = 20, I = 12 A, B = 0.80 T,
l = 10 cm = 10 \(\times\) 10^-2 m, \(\theta\) = 30°
\(\because\) Area, A = l² = (10 \(\times\)10^-2)² = 100 \(\times\)10^-4 m²
As, \(\tau\)= NBIA sin \(\theta\)
\(\Rightarrow\) \(\tau\)= 20 \(\times\)0.80 \(\times\) 12 \(\times\)100 \(\times\)10^-4 \(\times\)sin 30°
= 9600 \(\times\) 10^-4 = 0.96 N-m

(i) Here, the direction of magnetic field is perpendicularly inwards to the plane of paper. If a wire of irregular shape turns into a circular shape, then its area increases (\(\because \) the circular loop has greater area than the loop of irregular shape) so that, the magnetic flux linked also increases. Now, the induced current is produced in a direction such that it decreases the magnetic field [i.e. the current will flow in such a direction, so that the wire forming the loop is pulled inward in all directions (to decrease the area)], i.e current is in anti-clockwise direction, ie. along adcba.  
(ii) When a circular loop deforms into a narrow straight wire, the magnetic flux linked with it also decreases. The current induced due to change in flux will flow in such a direction that it will oppose the decrease in magnetic flux, so it will flow anti-clockwise, i.e along adcba due to which the magnetic field produced will be out of the plane of paper. 

F = Il x B
F = Il B sinθ
The force per unit length is
f = F / l = I B sinθ
(a) When the current is flowing from east to west,
θ = 90°
Hence,
f = I B
= 1 x 3x 10^–5 = 3 x 10^–5 N m^–1
This is larger than the value 2 x 10^–7 Nm^–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
θ = 0^o
f = 0
Hence there is no force on the conductor.

Given, slope of graph, tan \(\theta\) = 4.12 \(\times\) 10^-15 V-s

Charge on electron, e = 1.6 \(\times\)10^-19 C
Slope of graph of cut off voltage versus frequency is tan \(\theta=\frac{V}{v}\)
We know that, bv = eV or \(\frac{V}{v}=\frac{b}{e}\)
\(\begin{aligned} \therefore & \frac{h}{e}=4.12 \times 10^{-15} \end{aligned}\)
\(\begin{aligned} \Rightarrow & h=1.6 \times 10^{-19} \times 4.12 \times 10^{-15} \end{aligned}\)
\(\begin{aligned} =6.592 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned}\)

(a) Each photon has an energy
\(E=hv=6.63\times { 10 }^{ -34 }J s\times 6.0\times { 10 }^{ 14 }Hz\)
(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. Then
\(n=\frac { P }{ E } =\frac { 2.0\times { 10 }^{ -3 }W }{ 3.98\times { 10 }^{ -19 }J } \)
\(=5.0\times { 10 }^{ 15 }\) photons per second.

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

Number density of free electrons in a copper conductor, n = 8.5 x 10²⁸ m^-3 Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where,
e = Electric charge = 1.6 x 10⁻¹⁹ C
V_d = Drift velocity = \(\frac{\text { Length of the wire (l) }}{\text { Time taken to cover l(t) }}\)
\(I=n A e \frac{l}{t}\)
\(t=n A e \frac{l}{I}\)
\(=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(=2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 10⁴ s.

A) Step 1: Find the emf developed in the loop
Formula Used: \(\mathrm{e}=\mathrm{BIV}\)
Strength of magnetic field, \(B=0.3 \mathrm{~T}\)
Velocity of the loop, \(v=1 \mathrm{~cm} / \mathrm{s}=0.01 \mathrm{~m} / \mathrm{s}\)
emf developed in the loop is given as:
\( \mathrm{e}=\mathrm{Blv} \)
\(=0.3 \times 0.08 \times 0.01=2.4 \times 10^{-4} \mathrm{~V}\)
Step 2: Find the time taken to travel.
Formula Used: \(\mathrm{t}=\frac{\text { Distance travelled }}{\text { Velocity }}\)
Time taken to travel along the width, \(\mathrm{t}=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{\mathrm{bv}}{\mathrm{v}}\)
\(=\frac{0.02}{0.01}=2 \mathrm{~s}\)
Final answer : \(\mathrm{e}=2.4 \times 10^{-4} \mathrm{v}\)
\(\mathrm{t}=2 \mathrm{~s}\)
B) Step 1: Find the emf developed in the loop emf developed, e = BIv
\(e=0.3 \times 0.02 \times 0.01 \)
\(e=0.6 \times 10^{-4} V\)
Step 2: Find the time taken to travel.
Time taken to travel along the length, \(\mathrm{t}=\frac{\text { Distance travelle }}{\text { Velocity }}=\frac{\mathrm{d}}{\mathrm{v}}\)
\(\mathrm{t}=\frac{0.08}{0.01}=8 \mathrm{~s}\)
Hence, the induced voltage is \(0.6 \times 10^{-4} \mathrm{~V}\) which lasts for 8 s .
Final answer: \(\mathrm{e}=0.6 \times 10^{-4} \mathrm{~V}\)
\(\mathrm{t}=8 \mathrm{~s}\)

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge q =5 \(\mu\)C is placed. O is the centre of the hexagon.

Given, AB = BC = CD = DE
= EF = FA = d = 10 cm
As, the hexagon has six equilateral triangles, so the distance of centre O from every vertex is 10 cm.
i.e. OA = OB = OC = OD
= OE = OF = d = 10 cm
\(\therefore\) Potential at point O = Sum of potentials at centre O due to individual point charge
i.e. V_O = V_A + V_B + V_C + V_D + V_E + V_F
\(=\frac{1}{4\pi \varepsilon_{0}}.\left [ \frac{q}{OA}+\frac{q}{OB}+\frac{q}{OC}+\frac{q}{OD}+\frac{q}{OE}+\frac{q}{OF} \right ]\)
\(=\frac{1}{4\pi \varepsilon _{0}}.\frac{6q}{d}\)    \(\left [ \because V=\frac{1}{4\pi \varepsilon _{0}.\frac{q}{r}} \right ]\)
Putting the values, we get
\(=9\times 10^{9}\times \frac{6\times 5 \times 10^{-6}}{10\times 10^{-2}}\)
= 2.7 \(\times\)10⁶ V