Chapter 4: Moving Charges and Magnetism

For the wire to be suspended in mid-air, the upward magnetic force on the current-carrying wire must exactly balance its weight, which acts downward.

The weight of the wire is $$W = mg$$, and the magnetic force on a straight wire of length $$L$$ carrying a current $$I$$ in a field $$B$$ perpendicular to it is $$F = BIL$$.

For equilibrium the two are equal:

$$BIL = mg$$

$$B = \frac{mg}{IL}$$

Substituting $$m = 200 \, \mathrm{g} = 0.2 \, \mathrm{kg}$$, $$L = 1.5 \, \mathrm{m}$$, $$I = 2 \, \mathrm{A}$$ and $$g = 9.8 \, \mathrm{m/s^2}$$:

$$B = \frac{0.2 \times 9.8}{2 \times 1.5} = \frac{1.96}{3} \approx 0.65 \, \mathrm{T}$$

The field must be horizontal and perpendicular to the wire, directed so that the force $$I\,\mathbf{L}\times\mathbf{B}$$ points vertically upward.

The magnetic (Lorentz) force on a charge $$q$$ moving with velocity $$\mathbf{v}$$ in a field $$\mathbf{B}$$ is $$\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}$$.

Here $$\mathbf{v} = v\,\hat{\mathbf{i}}$$ (along $$+x$$) and $$\mathbf{B} = B\,\hat{\mathbf{j}}$$ (along $$+y$$). The vector product is

$$\mathbf{v}\times\mathbf{B} = (v\,\hat{\mathbf{i}})\times(B\,\hat{\mathbf{j}}) = vB\,(\hat{\mathbf{i}}\times\hat{\mathbf{j}}) = vB\,\hat{\mathbf{k}}$$

So $$\mathbf{v}\times\mathbf{B}$$ points along the positive $$z$$-axis.

(a) Electron (charge $$q = -e$$):

$$\mathbf{F} = (-e)(vB\,\hat{\mathbf{k}}) = -evB\,\hat{\mathbf{k}}$$

The force is directed along the negative $$z$$-axis.

(b) Proton (charge $$q = +e$$):

$$\mathbf{F} = (+e)(vB\,\hat{\mathbf{k}}) = evB\,\hat{\mathbf{k}}$$

The force is directed along the positive $$z$$-axis.

The two forces are equal in magnitude and opposite in direction, as expected for charges of opposite sign moving the same way.

Radius of the path. The magnetic force provides the centripetal force needed for the circular motion:

$$qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB}$$

Substituting $$m = 9\times10^{-31}\,\mathrm{kg}$$, $$v = 3\times10^{7}\,\mathrm{m/s}$$, $$q = 1.6\times10^{-19}\,\mathrm{C}$$ and $$B = 6\times10^{-4}\,\mathrm{T}$$:

$$r = \frac{(9\times10^{-31})(3\times10^{7})}{(1.6\times10^{-19})(6\times10^{-4})} = \frac{2.7\times10^{-23}}{9.6\times10^{-23}} \approx 0.28 \, \mathrm{m}$$

Frequency. The frequency of revolution is

$$\nu = \frac{v}{2\pi r} = \frac{qB}{2\pi m}$$

$$\nu = \frac{(1.6\times10^{-19})(6\times10^{-4})}{2\pi(9\times10^{-31})} = \frac{9.6\times10^{-23}}{5.655\times10^{-30}} \approx 1.7\times10^{7} \, \mathrm{Hz}$$

That is about $$17 \, \mathrm{MHz}$$.

Energy. The kinetic energy of the electron is

$$E = \frac{1}{2}mv^2 = \frac{1}{2}(9\times10^{-31})(3\times10^{7})^2 = \frac{1}{2}(9\times10^{-31})(9\times10^{14})$$

$$E = 4.05\times10^{-16} \, \mathrm{J}$$

Converting to electron-volts using $$1 \, \mathrm{eV} = 1.6\times10^{-19} \, \mathrm{J}$$:

$$E = \frac{4.05\times10^{-16}}{1.6\times10^{-19}} \, \mathrm{eV} \approx 2.5\times10^{3} \, \mathrm{eV} = 2.5 \, \mathrm{keV}$$

The Biot–Savart law gives the field of a small current element as

$$\Delta\mathbf{B} = \frac{\mu_0}{4\pi}\,\frac{I\,\Delta\mathbf{l}\times\hat{\mathbf{r}}}{r^2}$$

Here the element $$\Delta\mathbf{l} = \Delta x\,\hat{\mathbf{i}}$$ sits at the origin, and the field point lies on the $$y$$-axis, so the position vector is $$\mathbf{r} = r\,\hat{\mathbf{j}}$$ with $$r = 0.5 \, \mathrm{m}$$. The element points along $$x$$ while $$\mathbf{r}$$ points along $$y$$, so the angle between them is $$90°$$ and $$\sin\theta = 1$$.

Magnitude.

$$\Delta B = \frac{\mu_0}{4\pi}\,\frac{I\,\Delta x\,\sin\theta}{r^2} = \frac{\mu_0}{4\pi}\,\frac{I\,\Delta x}{r^2}$$

With $$\dfrac{\mu_0}{4\pi} = 10^{-7} \, \mathrm{T\,m/A}$$, $$I = 10 \, \mathrm{A}$$, $$\Delta x = 1 \, \mathrm{cm} = 10^{-2}\,\mathrm{m}$$ and $$r = 0.5 \, \mathrm{m}$$:

$$\Delta B = 10^{-7}\times\frac{10\times10^{-2}}{(0.5)^2} = 10^{-7}\times\frac{0.1}{0.25} = 4\times10^{-8} \, \mathrm{T}$$

Direction. $$\Delta\mathbf{l}\times\mathbf{r} = (\Delta x\,\hat{\mathbf{i}})\times(r\,\hat{\mathbf{j}}) = \Delta x\,r\,(\hat{\mathbf{i}}\times\hat{\mathbf{j}}) = \Delta x\,r\,\hat{\mathbf{k}}$$, so the field points along the positive $$z$$-axis.

(a) Field due to the straight segments. By the Biot–Savart law each current element contributes $$dB \propto I\,\mathbf{dl}\times\hat{\mathbf{r}}$$. For the two straight segments, every element lies along the straight line that passes through the centre of the arc, so $$\mathbf{dl}$$ and $$\hat{\mathbf{r}}$$ (the direction from the element to the centre) are either parallel or anti-parallel. Hence $$\mathbf{dl}\times\hat{\mathbf{r}} = 0$$ for every element.

Therefore the straight segments contribute nothing to the field at the centre: $$B_{\text{straight}} = 0$$.

(b) Field due to the semicircle. A full circular loop produces a field $$\mu_0 I/2R$$ at its centre. A semicircle is exactly half the loop, so it produces half this value:

$$B = \frac{1}{2}\cdot\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R}$$

$$B = \frac{(4\pi\times10^{-7})(12)}{4\times(0.02)} = \frac{1.508\times10^{-5}}{0.08} \approx 1.9\times10^{-4} \, \mathrm{T}$$

It differs from a full loop in magnitude — it is exactly half as large. It resembles the full loop in direction: the field is normal to the plane of the arc, its sense given by the right-hand rule (here, directed into the page).

(c) Arc bent the opposite way. The magnitude of the field is unchanged, $$B \approx 1.9\times10^{-4} \, \mathrm{T}$$, since the geometry of the semicircle is identical. However, the sense in which the current circulates around the arc is reversed, so the direction of $$\mathbf{B}$$ is reversed — it would now point out of the page.

The magnetic field at the centre of a tightly wound coil of $$N$$ turns, radius $$R$$, carrying a current $$I$$ is the field of a single loop multiplied by the number of turns:

$$B = \frac{\mu_0 N I}{2R}$$

Substituting $$N = 100$$, $$I = 1 \, \mathrm{A}$$, $$R = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$$ and $$\mu_0 = 4\pi\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = \frac{(4\pi\times10^{-7})(100)(1)}{2\times0.1} = \frac{1.2566\times10^{-4}}{0.2}$$

$$B \approx 6.28\times10^{-4} \, \mathrm{T}$$

Because the current is uniformly distributed and the wire has circular symmetry, the field $$\mathbf{B}$$ is tangential to circles drawn concentric with the wire's axis, and its magnitude is constant on each such circle. We therefore apply Ampère's circuital law on an Amperian loop that is a circle of radius $$r$$:

$$\oint\mathbf{B}\cdot\mathbf{dl} = \mu_0 I_{\text{enc}} \quad\Rightarrow\quad B\,(2\pi r) = \mu_0 I_{\text{enc}}$$

Region $$r > a$$ (outside the wire). The Amperian loop encloses the entire current $$I$$:

$$B\,(2\pi r) = \mu_0 I \quad\Rightarrow\quad B = \frac{\mu_0 I}{2\pi r}$$

The field falls off as $$1/r$$, exactly as for a thin straight wire.

Region $$r < a$$ (inside the wire). Since the current is uniform, the current enclosed by a circle of radius $$r$$ is the total current scaled by the ratio of areas:

$$I_{\text{enc}} = I\,\frac{\pi r^2}{\pi a^2} = I\,\frac{r^2}{a^2}$$

Ampère's law then gives

$$B\,(2\pi r) = \mu_0 I\,\frac{r^2}{a^2} \quad\Rightarrow\quad B = \frac{\mu_0 I r}{2\pi a^2}$$

Inside the wire the field grows linearly with $$r$$. At the surface $$r = a$$ both expressions give the same value, $$B = \dfrac{\mu_0 I}{2\pi a}$$, so the field is continuous there.

The number of turns per unit length of the solenoid is

$$n = \frac{N}{L} = \frac{500}{0.5} = 1000 \, \mathrm{turns/m}$$

The solenoid is long compared with its radius ($$L = 0.5 \, \mathrm{m}$$ versus $$r = 1 \, \mathrm{cm}$$), so we may treat it as an ideal solenoid, for which the field deep inside is uniform and given by

$$B = \mu_0 n I$$

Substituting $$n = 1000 \, \mathrm{m^{-1}}$$, $$I = 5 \, \mathrm{A}$$ and $$\mu_0 = 4\pi\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = (4\pi\times10^{-7})(1000)(5) = 4\pi\times10^{-7}\times5000$$

$$B \approx 6.28\times10^{-3} \, \mathrm{T}$$

The force per unit length on a straight conductor carrying current $$I$$ in a field $$B$$ is

$$f = \frac{F}{L} = I B\sin\theta$$

where $$\theta$$ is the angle between the current direction and $$\mathbf{B}$$. Here $$B = 3.0\times10^{-5} \, \mathrm{T}$$ points from geographic south to north, and $$I = 1 \, \mathrm{A}$$.

(a) Current from east to west. The current (east-to-west) is perpendicular to the field (south-to-north), so $$\theta = 90°$$:

$$f = I B \sin 90° = (1)(3.0\times10^{-5})(1) = 3.0\times10^{-5} \, \mathrm{N/m}$$

For the direction, $$\mathbf{F} = I\,\mathbf{L}\times\mathbf{B}$$: with the current pointing west and $$\mathbf{B}$$ pointing north, the cross product gives a force directed vertically downward.

(b) Current from south to north. Now the current is parallel to the field, so $$\theta = 0°$$:

$$f = I B \sin 0° = 0$$

There is no force on the conductor.

(a) Field at the centre of the coil.

$$B = \frac{\mu_0 N I}{2R} = \frac{(4\pi\times10^{-7})(100)(3.2)}{2\times0.10}$$

$$B = \frac{4.02\times10^{-4}}{0.20} \approx 2.0\times10^{-3} \, \mathrm{T}$$

(b) Magnetic moment of the coil. The moment is $$m = N I A$$, with $$A = \pi R^2$$:

$$m = N I \pi R^2 = (100)(3.2)\,\pi\,(0.10)^2 = 3.2\pi \approx 10 \, \mathrm{A\,m^2}$$

(c) Torques in the initial and final positions. The torque on the coil is $$\tau = mB\sin\theta$$, where $$\theta$$ is the angle between the magnetic moment $$\mathbf{m}$$ (along the coil's axis) and $$\mathbf{B}$$.

Initial position: the axis of the coil is along $$\mathbf{B}$$, so $$\theta = 0°$$ and

$$\tau_{\text{initial}} = mB\sin 0° = 0$$

Final position: the coil has turned through $$90°$$, so $$\theta = 90°$$ and

$$\tau_{\text{final}} = mB\sin 90° = (10)(2) = 20 \, \mathrm{N\,m}$$

(d) Angular speed after rotating $$90°$$. The work done by the magnetic torque as the coil turns from $$\theta = 0$$ to $$\theta = 90°$$ goes entirely into rotational kinetic energy:

$$W = \int_0^{\pi/2} \tau\,d\theta = \int_0^{\pi/2} mB\sin\theta\,d\theta = mB\big[-\cos\theta\big]_0^{\pi/2} = mB$$

Setting this equal to $$\tfrac{1}{2}I\omega^2$$:

$$mB = \frac{1}{2}I\omega^2 \quad\Rightarrow\quad \omega = \sqrt{\frac{2mB}{I}}$$

$$\omega = \sqrt{\frac{2\times20}{0.1}} = \sqrt{400} = 20 \, \mathrm{rad/s}$$

The circuit (Fig. 4.23) consists of a cell of emf $$\varepsilon = 3 \, \mathrm{V}$$ in series with a resistance $$R = 3 \, \Omega$$ and the measuring instrument. The current is $$I = \dfrac{\varepsilon}{R_{\text{total}}}$$, where $$R_{\text{total}}$$ must include the resistance of the instrument itself.

(a) The instrument is the bare galvanometer ($$R_G = 60.00 \, \Omega$$). Its large resistance adds to the circuit:

$$R_{\text{total}} = R + R_G = 3 + 60 = 63 \, \Omega$$

$$I = \frac{\varepsilon}{R + R_G} = \frac{3}{63} \approx 0.048 \, \mathrm{A}$$

The instrument badly disturbs the circuit — the measured current is far below the true value.

(b) Galvanometer converted to an ammeter with a shunt $$r_s = 0.02 \, \Omega$$. The ammeter's effective resistance is the parallel combination of $$R_G$$ and $$r_s$$:

$$R_A = \frac{R_G\,r_s}{R_G + r_s} = \frac{60\times0.02}{60 + 0.02} = \frac{1.2}{60.02} \approx 0.02 \, \Omega$$

$$R_{\text{total}} = R + R_A = 3 + 0.02 = 3.02 \, \Omega$$

$$I = \frac{\varepsilon}{R + R_A} = \frac{3}{3.02} \approx 0.99 \, \mathrm{A}$$

A properly shunted ammeter has a very small resistance and reads almost the true current.

(c) Ideal ammeter ($$R_A = 0$$):

$$I = \frac{\varepsilon}{R} = \frac{3}{3} = 1.00 \, \mathrm{A}$$

This is the true current in the circuit. Comparing the three cases shows clearly why an ammeter must have as small a resistance as possible.

The magnetic field at the centre of a circular coil of $$N$$ turns, each of radius $$R$$, carrying a current $$I$$ is

$$B = \frac{\mu_0 N I}{2R}$$

Here $$N = 100$$, $$R = 8.0 \, \mathrm{cm} = 0.080 \, \mathrm{m}$$, $$I = 0.40 \, \mathrm{A}$$ and $$\mu_0 = 4\pi\times10^{-7} \, \mathrm{T\,m/A}$$.

$$B = \frac{(4\pi\times10^{-7})(100)(0.40)}{2\times0.080} = \frac{5.03\times10^{-5}}{0.16}$$

$$B \approx 3.1\times10^{-4} \, \mathrm{T}$$

The field is directed along the axis of the coil, its sense given by the right-hand rule.

The magnetic field at a perpendicular distance $$r$$ from a long straight wire carrying a current $$I$$ is

$$B = \frac{\mu_0 I}{2\pi r}$$

Substituting $$I = 35 \, \mathrm{A}$$, $$r = 20 \, \mathrm{cm} = 0.20 \, \mathrm{m}$$ and $$\dfrac{\mu_0}{2\pi} = 2\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = \frac{(2\times10^{-7})(35)}{0.20} = \frac{7\times10^{-6}}{0.20}$$

$$B = 3.5\times10^{-5} \, \mathrm{T}$$

The magnitude of the field at a perpendicular distance $$r$$ from a long straight wire is

$$B = \frac{\mu_0 I}{2\pi r}$$

With $$I = 50 \, \mathrm{A}$$, $$r = 2.5 \, \mathrm{m}$$ and $$\dfrac{\mu_0}{2\pi} = 2\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = \frac{(2\times10^{-7})(50)}{2.5} = \frac{10^{-5}}{2.5} = 4\times10^{-6} \, \mathrm{T}$$

Direction. The current flows from north to south, and the field point lies to the east of the wire. Apply the right-hand rule: point the thumb of the right hand along the current (towards the south); the curled fingers then give the sense of $$\mathbf{B}$$ around the wire. On the eastern side of the wire the fingers point upward, so the magnetic field at this point is directed vertically upward.

The magnitude of the field at a perpendicular distance $$r$$ from a long straight wire is

$$B = \frac{\mu_0 I}{2\pi r}$$

With $$I = 90 \, \mathrm{A}$$, $$r = 1.5 \, \mathrm{m}$$ and $$\dfrac{\mu_0}{2\pi} = 2\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = \frac{(2\times10^{-7})(90)}{1.5} = \frac{1.8\times10^{-5}}{1.5} = 1.2\times10^{-5} \, \mathrm{T}$$

Direction. The current flows from east to west, and the field point is directly below the line. Apply the right-hand rule: point the thumb along the current (towards the west); the curled fingers give the sense of $$\mathbf{B}$$. At a point below the wire the fingers point towards the south, so the magnetic field there is directed towards the south.

The magnetic force per unit length on a current-carrying wire is

$$f = \frac{F}{L} = I B \sin\theta$$

where $$\theta$$ is the angle between the wire (the current direction) and the field $$\mathbf{B}$$.

Substituting $$I = 8 \, \mathrm{A}$$, $$B = 0.15 \, \mathrm{T}$$ and $$\theta = 30°$$:

$$f = (8)(0.15)\sin 30° = (8)(0.15)(0.5)$$

$$f = 0.6 \, \mathrm{N/m}$$

Inside a solenoid the magnetic field is uniform and directed along the axis. The wire is placed perpendicular to this axis, so the angle between the current direction and the field is $$\theta = 90°$$.

The magnetic force on a straight current-carrying wire is

$$F = B I L \sin\theta$$

Substituting $$B = 0.27 \, \mathrm{T}$$, $$I = 10 \, \mathrm{A}$$, $$L = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m}$$ and $$\theta = 90°$$:

$$F = (0.27)(10)(0.03)\sin 90° = (0.27)(10)(0.03)(1)$$

$$F = 8.1\times10^{-2} \, \mathrm{N}$$

The force is perpendicular both to the wire and to the axis of the solenoid.

Two long parallel wires carrying currents $$I_A$$ and $$I_B$$ and separated by a distance $$d$$ exert a force on each other. The force per unit length is

$$f = \frac{\mu_0 I_A I_B}{2\pi d}$$

Substituting $$I_A = 8.0 \, \mathrm{A}$$, $$I_B = 5.0 \, \mathrm{A}$$, $$d = 4.0 \, \mathrm{cm} = 0.04 \, \mathrm{m}$$ and $$\dfrac{\mu_0}{2\pi} = 2\times10^{-7} \, \mathrm{T\,m/A}$$:

$$f = \frac{(2\times10^{-7})(8.0)(5.0)}{0.04} = \frac{8.0\times10^{-6}}{0.04} = 2\times10^{-4} \, \mathrm{N/m}$$

The force on a section of length $$L = 10 \, \mathrm{cm} = 0.10 \, \mathrm{m}$$ of wire A is therefore

$$F = f\,L = (2\times10^{-4})(0.10) = 2\times10^{-5} \, \mathrm{N}$$

Since the two currents flow in the same direction, the force between the wires is attractive. (By Newton's third law, the force on an equal section of wire B is the same in magnitude.)

The total number of turns is the number of layers multiplied by the turns per layer:

$$N = 5\times400 = 2000 \, \mathrm{turns}$$

The number of turns per unit length is

$$n = \frac{N}{L} = \frac{2000}{0.80} = 2500 \, \mathrm{turns/m}$$

For a long solenoid the field near the centre is uniform and given by

$$B = \mu_0 n I$$

Substituting $$n = 2500 \, \mathrm{m^{-1}}$$, $$I = 8.0 \, \mathrm{A}$$ and $$\mu_0 = 4\pi\times10^{-7} \, \mathrm{T\,m/A}$$:

$$B = (4\pi\times10^{-7})(2500)(8.0)$$

$$B \approx 2.5\times10^{-2} \, \mathrm{T}$$

(The diameter $$1.8 \, \mathrm{cm}$$ is small compared with the length $$80 \, \mathrm{cm}$$, which justifies using the ideal long-solenoid formula.)

The torque on a current-carrying coil of $$N$$ turns and area $$A$$ placed in a uniform field $$B$$ is

$$\tau = N I A B \sin\theta$$

where $$\theta$$ is the angle between the normal to the plane of the coil and $$\mathbf{B}$$.

The coil is a square of side $$10 \, \mathrm{cm} = 0.10 \, \mathrm{m}$$, so its area is

$$A = (0.10)^2 = 0.01 \, \mathrm{m^2}$$

Substituting $$N = 20$$, $$I = 12 \, \mathrm{A}$$, $$A = 0.01 \, \mathrm{m^2}$$, $$B = 0.80 \, \mathrm{T}$$ and $$\theta = 30°$$:

$$\tau = (20)(12)(0.01)(0.80)\sin 30°$$

$$\tau = (20)(12)(0.01)(0.80)(0.5) = 0.96 \, \mathrm{N\,m}$$

For a moving-coil galvanometer with $$N$$ turns, coil area $$A$$, field $$B$$, coil resistance $$R$$ and spring (torsion) constant $$k$$, the equilibrium deflection $$\phi$$ satisfies $$k\phi = NIAB$$. Hence:

Current sensitivity (deflection per unit current):

$$\frac{\phi}{I} = \frac{N B A}{k}$$

Voltage sensitivity (deflection per unit voltage, with $$V = IR$$):

$$\frac{\phi}{V} = \frac{\phi}{IR} = \frac{N B A}{k R}$$

(a) Ratio of current sensitivities ($$\mathrm{M_2}$$ to $$\mathrm{M_1}$$). Since the spring constant $$k$$ is the same for both meters, it cancels:

$$\frac{(\phi/I)_2}{(\phi/I)_1} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1}$$

$$= \frac{(42)(0.50)(1.8\times10^{-3})}{(30)(0.25)(3.6\times10^{-3})} = \frac{0.0378}{0.0270} = 1.4$$

So the current sensitivity of $$\mathrm{M_2}$$ is $$1.4$$ times that of $$\mathrm{M_1}$$.

(b) Ratio of voltage sensitivities ($$\mathrm{M_2}$$ to $$\mathrm{M_1}$$):

$$\frac{(\phi/V)_2}{(\phi/V)_1} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1}\times\frac{R_1}{R_2} = 1.4\times\frac{10}{14}$$

$$= 1.4\times0.714 = 1.0$$

So the voltage sensitivities of the two meters are equal.

Why the path is a circle. The magnetic force on the moving electron is $$\mathbf{F} = -e\,\mathbf{v}\times\mathbf{B}$$. This force is always perpendicular to the velocity $$\mathbf{v}$$, so it does no work on the electron — the speed $$v$$ (and hence the kinetic energy) stays constant. A force of constant magnitude that always acts perpendicular to a constant-speed velocity is exactly a centripetal force: it continually changes the direction of $$\mathbf{v}$$ without changing its magnitude. Since the electron is shot in normal to $$\mathbf{B}$$, it stays in the plane perpendicular to $$\mathbf{B}$$ and moves in a circle at constant speed (uniform circular motion).

Radius of the orbit. The magnetic force supplies the centripetal force:

$$e v B = \frac{m_e v^2}{r} \quad\Rightarrow\quad r = \frac{m_e v}{e B}$$

Here $$B = 6.5 \, \mathrm{G} = 6.5\times10^{-4} \, \mathrm{T}$$ and $$v = 4.8\times10^{6} \, \mathrm{m\,s^{-1}}$$. Taking the electronic charge to be its standard value $$e = 1.6\times10^{-19} \, \mathrm{C}$$ and $$m_e = 9.1\times10^{-31} \, \mathrm{kg}$$:

$$r = \frac{(9.1\times10^{-31})(4.8\times10^{6})}{(1.6\times10^{-19})(6.5\times10^{-4})}$$

$$r = \frac{4.368\times10^{-24}}{1.04\times10^{-22}} \approx 4.2\times10^{-2} \, \mathrm{m}$$

So the radius of the circular orbit is about $$4.2 \, \mathrm{cm}$$. (The charge of the electron is $$1.6\times10^{-19} \, \mathrm{C}$$; the value $$1.5\times10^{-19} \, \mathrm{C}$$ quoted in the question is a misprint.)

For the uniform circular motion of the electron, the time taken for one revolution (the period) is

$$T = \frac{2\pi r}{v}$$

Using the radius found in Exercise 4.11, $$r = \dfrac{m_e v}{eB}$$, the period becomes

$$T = \frac{2\pi}{v}\cdot\frac{m_e v}{eB} = \frac{2\pi m_e}{eB}$$

The frequency of revolution is therefore

$$\nu = \frac{1}{T} = \frac{eB}{2\pi m_e}$$

Substituting $$e = 1.6\times10^{-19} \, \mathrm{C}$$, $$B = 6.5\times10^{-4} \, \mathrm{T}$$ and $$m_e = 9.1\times10^{-31} \, \mathrm{kg}$$:

$$\nu = \frac{(1.6\times10^{-19})(6.5\times10^{-4})}{2\pi(9.1\times10^{-31})} = \frac{1.04\times10^{-22}}{5.718\times10^{-30}}$$

$$\nu \approx 1.8\times10^{7} \, \mathrm{Hz} \;(\approx 18 \, \mathrm{MHz})$$

Does it depend on the speed? No. The final expression $$\nu = \dfrac{eB}{2\pi m_e}$$ contains only $$e$$, $$B$$ and $$m_e$$ — the speed $$v$$ has cancelled out. Physically, if the electron moves faster, its orbit radius grows in exact proportion ($$r\propto v$$), so the longer path is covered in the same time. Hence the frequency of revolution is independent of the electron's speed (and of the radius of the orbit). This speed-independent frequency is the cyclotron frequency.