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

Given, velocity of straight wire,
v = 5m/s
Horizontal component of the earth's magnetic field,

Length of the wire, \(\mathrm{I}=10 \mathrm{~m}\)
Falling speed of the wire, \(v=5.0 \mathrm{~m} / \mathrm{s}\)
Magnetic field strength, \(B=0.3 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
(a) Emf induced in the wire,
\( e=B l v \)
\(=0.3 \times 10^{-4} \times 5 \times 10 =1.5 \times 10^{-3} V\)
(b) Using Fleming's right-hand rule, it can be inferred that the direction of the induced emf is from West to East.
(c) The eastern end of the wire is at a higher potential.

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.

(a) To obtain a large deflection, one or more of the following steps can be taken:
(i) Use a rod made of soft iron inside the coil C₂ ,
(ii) Connect the coil to a powerful battery, and
(iii) Move the arrangement rapidly towards the test coil C₁ .
(b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current. In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary for his innovative skills.

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

Mutual inductance of a pair of coils, µ = 1.5 H
Initial current, I₁ = 0 A
Final current I₂ = 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where  is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:

Equating equations (1) and (2), we get

Hence, the change in the flux linkage is 30 Wb.

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

The angle \(\theta\) made by the area vector of the coil with the magnetic field is 45° . From Eq. (6.1), the initial magnetic flux is
\(\phi\) = BA cos \(\theta\)
\(\begin{aligned} =\frac{0.1 \times 10^{-2}}{\sqrt{2}} \mathrm{Wb} \end{aligned}\)
Final flux, F_min = 0
The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by
\(\varepsilon=\frac{\left|\Delta \Phi_B\right|}{\Delta t}=\frac{|(\Phi-0)|}{\Delta t}=\frac{10^{-3}}{\sqrt{2} \times 0.7}=1.0 \mathrm{mV}\)
And the magnitude of the current is
\(I=\frac{e}{R}=\frac{10^{-3} \mathrm{~V}}{0.5 \Omega}=2 \mathrm{~mA}\)
Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

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.

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: \(\mathrm{e}=0.6 \times 10^{-4} \mathrm{~V}\)
\(\mathrm{t}=8 \mathrm{~s}\)

Here, N = 30, R = 8.0 cm = 8 \(\times\)10^-2 m,
I = 6.0 A, \(\theta\)= 60° and B = 1.0 T
(i) The magnitude of the counter torque
= magnitude of the deflecting torque
= NAIB sin \(\theta\) = N .(\(\pi\)R²)IB sin \(\theta\)
= 30 \(\times\) 3.14 \(\times\) (8 \(\times\) 10^-2)² \(\times\) 6.0 \(\times\) 1.0 \(\times\) sin 60°
= 3.14 N.m
(ii) The answer would not change as area enclosed by the coil as well as all other particulars remain unaltered and the formula, \(\tau\) = NAIB sin \(\theta\) is true for planar coil for any shape.

Initial flux through the coil,
Φ_{B (initial)} = BA cos θ
= 3.0 x 10^–5 x (π x 10^–2) x cos 0º
= 3π x 10^–7 Wb
Final flux after the rotation,
Φ_{B (final)} = 3.0 x 10^–5 x (π x 10^–2) x cos 180°
= –3π x 10^–7 Wb
Therefore, estimated value of the induced emf is,
\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)
= 500 x (6π x 10^–7)/0.25
= 3.8 x 10^–3 V
I = ε/R = 1.9 x 10^–3 A
Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 x 10^-4 m²
Current in the solenoid, I = 4.0 A
(a) The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 x 4 x 1.6 x 10^-4
= 1.28 Am²
(b) Magnetic field, B = 7.5 x 10^-2
The angle between the magnetic field and the axis of the solenoid, θ = 30°
Torque, \(\tau\) = MB sinθ
=1.28 x 7.5 x 10^-2 sin30°
= 0.048J
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 0.048 J

(a) Magnetic moment, M = 1.5 J T^-1
Magnetic field strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic field, θ₁​ = 0°
Final angle between the axis and the magnetic field, θ₂ ​= 90°
The work required to make the magnetic moment normal to the direction of the magnetic field is given as:
W = -MB (cosθ₂​ - cosθ_1​)
= -1.5 x 0.22(cos90°- cos0°)
= - 0.33 (0 - 1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, θ₁ ​= 0°
Final angle between the axis and the magnetic field, θ₂ ​= 180°
The work required to make the magnetic moment opposite to the direction of the magnetic field is given as:
W = - MB (cosθ₂ ​- cosθ₁​)
= -1.5 x 0.22 (cos180° - cos0°)
= - 0.33 (- 1 - 1)
= 0.66 J
(b) For case (i):
θ = θ₂ ​= 90°
∴ Torque, \(\tau\) =MB sinθ
= MB sin 90°
= 0.33 J
The torque tends to align the magnitude moment vector along B.
For case (ii):
θ = θ₂ ​= 180° ∴ Torque, \(\tau\) = MB sinθ
= MB sin180°
= 0

(a) The field H is dependent of the material of the core, and is
H = nI = 1000 x 2.0 = 2 x 10³ A/m.
(b) The magnetic field B is given by
B = μ_r μ₀ H
= 400 x 4π x 10^-7 (N/A²) x 2 x 10³ (A/m)
= 1.0 T
(c) Magnetisation is given by
M = (B - μ₀ H)/ μ₀
= (μ_r μ₀ H - μ₀ H) / μ₀ = (μ_r – 1)H = 399 x H
\(\cong\)8 \(\times\)10⁵ A/m
d) The magnetising current I_M is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ_r n (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794 A.

Magnetic field strength, B = 6.5 G = 6.5 x 10^–4 T
Speed of the electron, v = 4.8 x 10⁶ m/s
Charge on the electron, e = 1.6 x 10^–19 C
Mass of the electron, m_e = 9.1 x 10^–31 kg
Angle between the shot electron and magnetic field, θ = 90°
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB sinθ
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
\(F_{c}=\frac{m v^{2}}{r}\)
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
F_C = F
\(\frac{m v^{2}}{r}=e v B \sin \theta\)
\(r=\frac{m v}{B e \sin \theta}\)
\(=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{6.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times \sin 90^{\circ}}\)
= 4.2 x 10 ^-2 m = 4.2 cm
Hence, the radius of the circular orbit of the electron is 4.2 cm.