Moving Charges and Magnetism

Given:
- Number of turns, $N = 100$
- Radius of coil, $R = 8.0\,\text{cm} = 0.08\,\text{m}$
- Current, $I = 0.40\,\text{A}$
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
The magnetic field at the centre of a circular coil of $N$ turns is:
$$B = \frac{\mu_0 N I}{2R}$$

Calculation:
$$B = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08}$$
$$B = \frac{4\pi \times 10^{-7} \times 40}{0.16}$$
$$B = \frac{4\pi \times 10^{-7} \times 250}{1}$$
$$B = 4\pi \times 250 \times 10^{-7}$$
$$B = 3.14 \times 10^{-4}\,\text{T}$$

Answer: The magnitude of the magnetic field at the centre of the coil is $B \approx 3.14 \times 10^{-4}\,\text{T}$.

Given:
- Current, $I = 35\,\text{A}$
- Distance from wire, $R = 20\,\text{cm} = 0.20\,\text{m}$
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
The magnetic field at a perpendicular distance $R$ from a long straight wire is:
$$B = \frac{\mu_0 I}{2\pi R}$$

Calculation:
$$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.20}$$
$$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.20}$$
$$B = \frac{2 \times 10^{-7} \times 35}{0.20}$$
$$B = \frac{70 \times 10^{-7}}{0.20}$$
$$B = 3.5 \times 10^{-5}\,\text{T}$$

Answer: The magnitude of the magnetic field at a point 20 cm from the wire is $B = 3.5 \times 10^{-5}\,\text{T}$.

Given:
- Current, $I = 50\,\text{A}$ (flowing from North to South)
- Distance from wire, $R = 2.5\,\text{m}$ (point is to the East)
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
$$B = \frac{\mu_0 I}{2\pi R}$$

Calculation:
$$B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5}$$
$$B = \frac{2 \times 10^{-7} \times 50}{2.5}$$
$$B = \frac{100 \times 10^{-7}}{2.5} = 4 \times 10^{-6}\,\text{T}$$

Direction:
Using the right-hand thumb rule: Point the thumb of the right hand in the direction of current (North to South, i.e., $-\hat{y}$ direction). The fingers curl from East (where the point is) in the downward direction.

Alternatively, using $\mathbf{B} = \frac{\mu_0 I}{2\pi R}\hat{\phi}$: The current is in the $-\hat{j}$ (South) direction, the point is in the $+\hat{i}$ (East) direction. The field direction is $(-\hat{j}) \times (+\hat{i}) = -(-\hat{k}) = +\hat{k}$...

More carefully: $\hat{l} \times \hat{r} = (-\hat{j}) \times (+\hat{i}) = -(\hat{j} \times \hat{i}) = -(-\hat{k}) = +\hat{k}$, i.e., vertically upward.

Answer: The magnitude of the magnetic field is $B = 4 \times 10^{-6}\,\text{T}$ and it is directed vertically upward (out of the horizontal plane).

Given:
- Current, $I = 90\,\text{A}$ (flowing East to West)
- Distance below the line, $R = 1.5\,\text{m}$
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
$$B = \frac{\mu_0 I}{2\pi R}$$

Calculation:
$$B = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5}$$
$$B = \frac{2 \times 10^{-7} \times 90}{1.5}$$
$$B = \frac{180 \times 10^{-7}}{1.5} = 1.2 \times 10^{-5}\,\text{T}$$

Direction:
The current flows from East to West (i.e., in the $-\hat{i}$ direction). The point is 1.5 m below the wire (i.e., in the $-\hat{k}$ direction).

Using the right-hand thumb rule: Point the thumb in the direction of current (East to West). The field lines form circles around the wire. At a point directly below the wire, the field is directed towards the South (i.e., $-\hat{j}$ direction).

Verification: $\hat{l} \times \hat{r} = (-\hat{i}) \times (-\hat{k}) = +(\hat{i} \times \hat{k}) = -\hat{j}$ (South). ✓

Answer: The magnitude of the magnetic field is $B = 1.2 \times 10^{-5}\,\text{T}$ and it is directed towards the South.

Given:
- Current, $I = 8\,\text{A}$
- Angle between wire and magnetic field, $\theta = 30°$
- Magnetic field, $B = 0.15\,\text{T}$

Formula used:
The force on a current-carrying conductor is $F = BIl\sin\theta$.
Therefore, force per unit length is:
$$\frac{F}{l} = BI\sin\theta$$

Calculation:
$$\frac{F}{l} = 0.15 \times 8 \times \sin 30°$$
$$\frac{F}{l} = 0.15 \times 8 \times 0.5$$
$$\frac{F}{l} = 0.6\,\text{N m}^{-1}$$

Answer: The magnitude of the magnetic force per unit length on the wire is $0.6\,\text{N m}^{-1}$.

Given:
- Length of wire, $l = 3.0\,\text{cm} = 0.03\,\text{m}$
- Current, $I = 10\,\text{A}$
- Magnetic field inside solenoid, $B = 0.27\,\text{T}$
- Angle between wire and field, $\theta = 90°$ (wire is perpendicular to the axis, and the field is along the axis)

Formula used:
$$F = BIl\sin\theta$$

Calculation:
$$F = 0.27 \times 10 \times 0.03 \times \sin 90°$$
$$F = 0.27 \times 10 \times 0.03 \times 1$$
$$F = 8.1 \times 10^{-2}\,\text{N}$$

Answer: The magnetic force on the wire is $F = 8.1 \times 10^{-2}\,\text{N}$.

Given:
- Current in wire A, $I_A = 8.0\,\text{A}$
- Current in wire B, $I_B = 5.0\,\text{A}$
- Separation between wires, $d = 4.0\,\text{cm} = 0.04\,\text{m}$
- Length of section considered, $l = 10\,\text{cm} = 0.10\,\text{m}$
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Formula used:
The force per unit length between two parallel current-carrying wires is:
$$\frac{F}{l} = \frac{\mu_0 I_A I_B}{2\pi d}$$

So the force on a length $l$ of wire A is:
$$F = \frac{\mu_0 I_A I_B l}{2\pi d}$$

Calculation:
$$F = \frac{4\pi \times 10^{-7} \times 8.0 \times 5.0 \times 0.10}{2\pi \times 0.04}$$
$$F = \frac{4\pi \times 10^{-7} \times 4.0}{2\pi \times 0.04}$$
$$F = \frac{2 \times 10^{-7} \times 8 \times 5 \times 0.10}{0.04}$$
$$F = \frac{2 \times 10^{-7} \times 4}{0.04}$$
$$F = \frac{8 \times 10^{-7}}{0.04} = 2 \times 10^{-5}\,\text{N}$$

Nature of force: Since the currents are in the same direction, the force is attractive.

Answer: The force on a 10 cm section of wire A is $F = 2 \times 10^{-5}\,\text{N}$, and it is attractive (directed towards wire B).

Given:
- Length of solenoid, $L = 80\,\text{cm} = 0.80\,\text{m}$
- Number of layers = 5
- Turns per layer = 400
- Total number of turns, $N = 5 \times 400 = 2000$
- Current, $I = 8.0\,\text{A}$
- $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Number of turns per unit length:
$$n = \frac{N}{L} = \frac{2000}{0.80} = 2500\,\text{turns/m}$$

Formula used:
The magnetic field inside a solenoid near its centre:
$$B = \mu_0 n I$$

Calculation:
$$B = 4\pi \times 10^{-7} \times 2500 \times 8.0$$
$$B = 4\pi \times 10^{-7} \times 20000$$
$$B = 4\pi \times 2 \times 10^{-3}$$
$$B = 8\pi \times 10^{-3}$$
$$B \approx 2.512 \times 10^{-2}\,\text{T}$$
$$B \approx 0.025\,\text{T}$$

(Note: The diameter of the solenoid is not needed for this calculation as the field inside an ideal solenoid does not depend on its cross-sectional dimensions.)

Answer: The magnitude of the magnetic field inside the solenoid near its centre is $B \approx 2.5 \times 10^{-2}\,\text{T}$.

Given:
- Side of square coil, $a = 10\,\text{cm} = 0.10\,\text{m}$
- Number of turns, $N = 20$
- Current, $I = 12\,\text{A}$
- Angle between normal to coil and magnetic field, $\theta = 30°$
- Magnetic field, $B = 0.80\,\text{T}$

Area of the coil:
$$A = a^2 = (0.10)^2 = 0.01\,\text{m}^2$$

Magnetic moment:
$$m = NIA = 20 \times 12 \times 0.01 = 2.4\,\text{A m}^2$$

Formula used:
The torque on a current-carrying coil in a magnetic field:
$$\tau = NIA B\sin\theta = mB\sin\theta$$

Calculation:
$$\tau = 2.4 \times 0.80 \times \sin 30°$$
$$\tau = 2.4 \times 0.80 \times 0.5$$
$$\tau = 0.96\,\text{N m}$$

Answer: The magnitude of the torque experienced by the coil is $\tau = 0.96\,\text{N m}$.

Given:
- $R_1 = 10\,\Omega,\; N_1 = 30,\; A_1 = 3.6 \times 10^{-3}\,\text{m}^2,\; B_1 = 0.25\,\text{T}$
- $R_2 = 14\,\Omega,\; N_2 = 42,\; A_2 = 1.8 \times 10^{-3}\,\text{m}^2,\; B_2 = 0.50\,\text{T}$
- Spring constants: $k_1 = k_2 = k$ (identical)

Concept:
For a moving coil galvanometer, the deflection is given by $k\phi = NIAB$.

- Current sensitivity $= \dfrac{\phi}{I} = \dfrac{NAB}{k}$
- Voltage sensitivity $= \dfrac{\phi}{V} = \dfrac{NAB}{kR}$

(a) Ratio of current sensitivity:
$$\frac{(I_s)_2}{(I_s)_1} = \frac{N_2 A_2 B_2 / k}{N_1 A_1 B_1 / k} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1}$$
$$= \frac{42 \times 1.8 \times 10^{-3} \times 0.50}{30 \times 3.6 \times 10^{-3} \times 0.25}$$
$$= \frac{42 \times 1.8 \times 0.50}{30 \times 3.6 \times 0.25}$$
$$= \frac{37.8}{27.0} = 1.4$$

$$\boxed{\frac{(I_s)_2}{(I_s)_1} = 1.4}$$

(b) Ratio of voltage sensitivity:
$$\frac{(V_s)_2}{(V_s)_1} = \frac{N_2 A_2 B_2 / (k R_2)}{N_1 A_1 B_1 / (k R_1)} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1} \times \frac{R_1}{R_2}$$
$$= 1.4 \times \frac{10}{14} = 1.4 \times \frac{5}{7} = 1.0$$

$$\boxed{\frac{(V_s)_2}{(V_s)_1} = 1.0}$$

Answer:
(a) The ratio of current sensitivity of M₂ to M₁ is 1.4.
(b) The ratio of voltage sensitivity of M₂ to M₁ is 1.0 (they are equal).

Given:
- Magnetic field, $B = 6.5\,\text{G} = 6.5 \times 10^{-4}\,\text{T}$
- Speed of electron, $v = 4.8 \times 10^6\,\text{m s}^{-1}$
- Charge of electron, $e = 1.6 \times 10^{-19}\,\text{C}$
- Mass of electron, $m_e = 9.1 \times 10^{-31}\,\text{kg}$
- The velocity is normal to the field.

Why the path is circular:
The magnetic force on the electron is $F = evB\sin\theta$. Since $v \perp B$, $\theta = 90°$, so $F = evB$. This force is always perpendicular to the velocity of the electron (it is a centripetal force). Since the force does no work (it is always perpendicular to displacement), the speed remains constant. The direction of force changes continuously, always pointing towards the centre of the circular path. Hence, the electron moves in a circular orbit.

Radius of circular orbit:
The magnetic force provides the centripetal force:
$$evB = \frac{m_e v^2}{r}$$
$$r = \frac{m_e v}{eB}$$

Calculation:
$$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$$
$$r = \frac{9.1 \times 4.8 \times 10^{-25}}{1.6 \times 6.5 \times 10^{-23}}$$
$$r = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}}$$
$$r = \frac{43.68}{10.4} \times 10^{-2}$$
$$r = 4.2 \times 10^{-2}\,\text{m} = 4.2\,\text{cm}$$

Answer: The path of the electron is circular because the magnetic force is always perpendicular to its velocity. The radius of the circular orbit is $r \approx 4.2\,\text{cm}$.

Given (from Exercise 4.11):
- $B = 6.5 \times 10^{-4}\,\text{T}$
- $e = 1.6 \times 10^{-19}\,\text{C}$
- $m_e = 9.1 \times 10^{-31}\,\text{kg}$

Formula used:
The cyclotron frequency (frequency of revolution) is:
$$\nu_c = \frac{eB}{2\pi m_e}$$

Calculation:
$$\nu_c = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}}$$
$$\nu_c = \frac{1.04 \times 10^{-22}}{5.7196 \times 10^{-30}}$$
$$\nu_c = \frac{1.04}{5.7196} \times 10^{8}$$
$$\nu_c \approx 1.82 \times 10^{7}\,\text{Hz} \approx 18\,\text{MHz}$$

Does the frequency depend on speed?
No, the frequency does not depend on the speed of the electron. This is because:
- The radius $r = \dfrac{m_e v}{eB}$ is proportional to $v$.
- The time period $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m_e}{eB}$, which is independent of $v$.
- Therefore, $\nu_c = \dfrac{1}{T} = \dfrac{eB}{2\pi m_e}$ is independent of the speed $v$.

This is the fundamental principle exploited in the cyclotron.

Answer: The frequency of revolution of the electron is $\nu_c \approx 1.82 \times 10^7\,\text{Hz}$. The frequency does not depend on the speed of the electron.

Part (a):

Given:
- Number of turns, $N = 30$
- Radius of coil, $R = 8.0\,\text{cm} = 0.08\,\text{m}$
- Current, $I = 6.0\,\text{A}$
- Magnetic field, $B = 1.0\,\text{T}$
- Angle between field lines and normal to coil, $\theta = 60°$

Area of the coil:
$$A = \pi R^2 = \pi \times (0.08)^2 = \pi \times 6.4 \times 10^{-3} = 2.011 \times 10^{-2}\,\text{m}^2$$

Torque on the coil:
$$\tau = NIAB\sin\theta$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times 1.0 \times \sin 60°$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times \frac{\sqrt{3}}{2}$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times 0.866$$
$$\tau = 180 \times 2.011 \times 10^{-2} \times 0.866$$
$$\tau = 180 \times 1.7415 \times 10^{-2}$$
$$\tau \approx 3.13\,\text{N m}$$

To prevent the coil from turning, the counter torque must be equal in magnitude:
$$\tau_{\text{counter}} \approx 3.13\,\text{N m}$$

Part (b):

No, the answer would not change. The torque on a planar current loop depends only on the magnetic moment $\mathbf{m} = NIA$, the magnetic field $B$, and the angle $\theta$ between $\mathbf{m}$ and $\mathbf{B}$. It does not depend on the shape of the loop, only on the area enclosed. Since the irregular planar coil encloses the same area $A$, carries the same current $I$, has the same number of turns $N$, and is placed in the same field at the same angle, the torque (and hence the required counter torque) remains the same at $\approx 3.13\,\text{N m}$.

Answer:
(a) The magnitude of the counter torque required is $\tau \approx 3.13\,\text{N m}$.
(b) No, the answer does not change, because torque depends only on the area of the loop, not its shape.