Here, the angle between the magnetic field and the direction of flow of current is 90°. Because the magnetic field due to a solenoid is along the axis of the solenoid and the wire is placed perpendicular to the axis.

Given, l = 3 cm = 3 \(\times\)10^-2 m
I = 10 A, B = 0.27 T
The magnitude of magnetic force on the wire,
F = IlB sin 90°
=10 \(\times\)3 \(\times\)10^-2 \(\times\)0.27 \(\times\)sin 90°
= 8.1 \(\times\)10^-2 N
According to right hand palm rule, the direction of magnetic force is perpendicular to plane of paper inwards.

Here, I = 8 A, \(\theta\) =30°, B = 0.15T, F = ?, l = 1 m
We know that, F = BIl sin \(\theta\)
\(\begin{aligned} & \frac{F}{l}=B I \sin \theta \end{aligned}\)
\(\begin{aligned} \frac{F}{l}=0.15 \times 8 \times \sin 30^{\circ} \end{aligned}\)
= 0.15 \(\times\)8 \(\times\)(1 / 2)
= 0.6 Nm^-1

Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
\(B=\frac{\mu_{0} 2 I}{4 \pi r}\)
Where,
μ_0 = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 90}{4 \pi \times 1.5}=1.2 \times 10^{-5} T\)
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
∴  Magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, B \(=\frac{\mu_{0} 2 I}{4 \pi r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 50}{4 \pi \times 2.5}\)
= 4 x 10 ^-6 T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Current in the wire, I = 35 A
Distance of a point from the wire, r = 20 cm = 0.2 m
Magnitude of the magnetic field at this point is given as:
\(B=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2}\)
= 3.3 x 10^-5 T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10^–5 T.

(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) From Eq.
\(B=\frac{\mu_{0} N I}{2 R}\)
Here, N = 100; I = 3.2 A, and R = 0.1 m. Hence,
\(B=\frac{4 \pi \times 10^{-7} \times 10^{2} \times 3.2}{2 \times 10^{-1}}=\frac{4 \times 10^{-5} \times 10}{2 \times 10^{-1}} \)   (using π x 3.2 = 10)
= 2 x 10^–3 T
The direction is given by the right-hand thumb rule.
(b) The magnetic moment is given by Eq.
m = N I A = N I π r² = 100 x 3.2 x 3.14 x 10^–2 = 10 A m²
The direction is once again given by the right hand thumb rule.
(c) τ = |m × B|    
= m B sinθ
Initially, θ = 0. Thus, initial torque τ i = 0. Finally, θ = π/2 (or 90º).
Thus, final torque τ_f = m B = 10 x 2 = 20 N m.
(d) From Newton’s second law,
\(\mathscr{I} \frac{\mathrm{d} \omega}{\mathrm{d} t}=m B \sin \theta\)
where \(\mathscr{I}\) is the moment of inertia of the coil. From chain rule,
\(\frac{\mathrm{d} \omega}{\mathrm{d} t}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \omega\)
Using this,
\(\mathscr{I} \omega \mathrm{d} \omega=m B \sin \theta \mathrm{d} \theta\)
Integrating from θ = 0 to θ = π / 2,
\(\mathscr{I} \int_{0}^{\omega_{f}} \omega \mathrm{d} \omega=m B \int_{0}^{\pi / 2} \sin \theta \mathrm{d} \theta\)
\(\mathscr{I}\frac{\omega_{f}^{2}}{2}=-\left.m B \cos \theta\right|_{0} ^{\pi / 2}=m B\)
\(\omega_{f}=\left(\frac{2 m B}{g}\right)^{1 / 2}=\left(\frac{2 \times 20}{10^{-1}}\right)^{1 / 2}=20 \mathrm{~s}^{-1}\)

(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).

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.

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.

Magnetic field strength, B = 6.5 x 10⁻⁴ T
Charge of the electron, e = 1.6 x 10⁻¹⁹ C
Mass of the electron, m_e = 9.1 x 10⁻³¹ kg
Velocity of the electron, v = 4.8 x 10⁶ m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = ν
Angular frequency of the electron = ω = 2πν
Velocity of the electron is related to the angular frequency as:
v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
\(e v B=\frac{m v^{2}}{r}\)
\(e B=\frac{m}{r}(r \omega)=\frac{m}{r}(2 \pi r v)\)
\(v=\frac{B e}{2 \pi m}\)
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
\(V=\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\)
= 18.2 x 10⁶ Hz
\(\approx\) 18 MHz
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Given, R₁ = 10 \(\Omega\), N₁ = 30, A₁ = 3.6 \(\times\)10^-3m²,
B₁ = 0.25 T, R₂ = 14 \(\Omega\), N₂ = 42.
A₂ = 1.8 \(\times\)10^-3 m², B₂ = 0.50 T
k₁ = k₂ (spring constants are smae)    ...(i)
(i) Using the formula of current sensitivity, \(I=\frac{N A B}{k}\)
\(\therefore \quad \frac{I_{S_2}}{I_{S_1}}=\frac{N_2 B_2 A_2 k_1}{N_1 B_1 A_1 k_2}=\frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}}\)
= 1.4     [from Eq. (i)]
(ii) Using the formula of voltage sensitivity,
\(\begin{aligned} V & =\frac{N A B}{k R} \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{V_{S_2}}{V_{S_1}} & =\frac{N_2 B_2 A_2 k_1 R_1}{k_2 R_2 N_1 B_1 A_1} \\ \end{aligned}\)
\(\begin{aligned} =\frac{42 \times 0.50 \times 1.8 \times 10^{-3} \times 10}{14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}} \end{aligned}\)
= 1             [from Eq. (i)]

Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
\(B=\frac{\mu_{0} N I}{l}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}\)
\(=8 \pi \times 10^{-3}=2.512 \times 10^{-2} T\)
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10^–2 T.

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.

Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, 
\(B=\frac{\mu_{0} N I}{2 R}=\frac{4 \pi \times 10^{-7} \times 10^{2} \times 1}{2 \times 10^{-1}}=2 \pi \times 10^{-4}=6.28 \times 10^{-4} \mathrm{~T}\) 

Using Eq. we find
r = m v / (qB) = 9 x 10^–31 kg x 3 x 10⁷ m s^–1 / ( 1.6 x 10^–19 C x 6 x 10^–4 T )
= 28 x 10^–2 m = 28 cm
ν = v / (2 \(\pi\)r) = 17 x 10⁶ s^–1 = 17 x 10⁶ Hz = 17 MHz.
E = (½ ) mv² = (½ ) 9 x 10^–31 kg x 9 x 10¹⁴ m²/s² = 40.5 x 10^–17 J
\(\approx \) 4 x 10^–16 J = 2.5 keV.

The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v x B is along the z-axis (screw rule or right-hand thumb rule). So,
(a) for electron it will be along –z axis.
(b) for a positive charge (proton) the force is along + z axis.

Here, n = 100, r = 8 cm = 8 \(\times\)10^-2m and I = 0.40 A
\(\therefore\) Magnetic field B at the centre,
\(\begin{aligned} B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I n}{r} & =\frac{10^{-7} \times 2 \times 3.14 \times 0.40 \times 100}{8 \times 10^{-2}} \\ \end{aligned}\)
\(\begin{aligned} =3.1 \times 10^{-4} \mathrm{~T} \end{aligned}\)

From Equation we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity.
m g = I l B
\(B=\frac{m g}{I l}\)
\(=\frac{0.2 \times 9.8}{2 \times 1.5}=0.65 \mathrm{~T}\)
Note that it would have been sufficient to specify m / l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 x 10^–5 T and we have ignored it.

Given, total number of turns, N = 500
Length of solenoid, l = 0.5 m
Current, I = 5 A
Radius, r = 1 cm = 10^-2 m
Here, \(\begin{aligned} \frac{l}{r} & =\frac{0.5}{10^{-2}}=50 \Rightarrow l>>r \\ \end{aligned}\)
\(\begin{aligned} \therefore B & =\mu_0 n I=\frac{\mu_0 N I}{l} \\ \end{aligned}\)
\(\begin{aligned} =4 \pi \times 10^{-7} \times \frac{500}{0.5} \times 5 \end{aligned}\)
= 6.28 \(\times\) 10^-3 T

\(|\mathrm{dB}|=\frac{\mu_{0}}{4 \pi} \frac{I \mathrm{~d} l \sin \theta}{r^{2}}\)
dl = Δx = 10⁻²m , I = 10 A, r = 0.5 m = y, \(\mu_{0} / 4 \pi=10^{-7} \frac{\mathrm{T} \mathrm{m}}{\mathrm{A}}\) θ = 90° ; sin θ = 1
\(|\mathrm{dB}|=\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}=4 \times 10^{-8} \mathrm{~T}\)
The direction of the field is in the +z-direction. This is so since
\(\mathrm{d} \mathbf{l} \times \mathbf{r}=\Delta x \hat{\mathbf{i}} \times y \hat{\mathbf{j}}=y \Delta x(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=y \Delta x \hat{\mathbf{k}}\)
We remind you of the following cyclic property of cross-products
\(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} ; \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}} ; \hat{\mathbf{k}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}\)
Note that the field is small in magnitude.

(a) No, because that would require \(\tau \) to be in the vertical direction. But \(\tau =IA\times B\), and since A of the horizontal loop is in the vertical direction, \(\tau \) would be in the plane of the loop for any B.
(b) Orientation of stable equilibrium is one where the are vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.
(c) It assumes circular shape with its plane normal to the field to maximize flux, since, for a given perimeter, a circle encloses greater areas than any other shape.