Chapter 4: Moving Charges and Magnetism (Magnetism)

$$B=\frac{{\mu }_{0}NI}{2R}=\frac{4\pi \times 10^{-7}\times 100\times 0.40}{2\times 0.08}\approx 3.1\times 10^{-4}\text{T}$$

$B\approx 3.1\times 10^{-4}$ T.

$$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 35}{0.20}=3.5\times 10^{-5}\text{T}$$

$B=3.5\times 10^{-5}$ T.

$$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 50}{2.5}=4\times 10^{-6}\text{T}$$

Direction (right-hand thumb rule): point the thumb along the current (towards south). The fingers curl around the wire; on the east side of the wire they point vertically upward.

$4\times 10^{-6}$ T, vertically upward.

$$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 90}{1.5}=1.2\times 10^{-5}\text{T}$$

Direction: thumb along the current (towards west); the curled fingers just below the wire point towards the south.

$1.2\times 10^{-5}$ T, towards the south.

$$\frac{F}{l}=BI\sin \theta =0.15\times 8\times \sin 30^{\circ }=0.6\text{N/m}$$

0.6 N m⁻¹.

The field inside a solenoid is along its axis, so the wire is perpendicular to $B$ ($\theta =90^{\circ }$):

$$F=BIl=0.27\times 10\times 0.03=8.1\times 10^{-2}\text{N}$$

The force is perpendicular to both the wire and the axis (direction by Fleming's left-hand rule).

$8.1\times 10^{-2}$ N, perpendicular to the wire and the solenoid's axis.

$$F=\frac{{\mu }_0{I}_A{I}_B}{2\pi d}l=\frac{2\times 10^{-7}\times 8.0\times 5.0}{0.04}\times 0.10=2\times 10^{-5}\text{N}$$

Currents in the same direction attract, so the force on A is towards B.

$2\times 10^{-5}$ N, attractive (towards B).

Total turns $N=5\times 400=2000$; turns per metre $n=\frac{2000}{0.80}=2500$ m⁻¹.

$$B={\mu }_{0}nI=4\pi \times 10^{-7}\times 2500\times 8.0\approx 2.5\times 10^{-2}\text{T}$$

The diameter does not enter, because the solenoid is long compared with its diameter.

$B\approx 2.5\times 10^{-2}$ T.

$A=(0.10{)}^2=0.01$ m².

$$\tau =NIAB\sin \theta =20\times 12\times 0.01\times 0.80\times \sin 30^{\circ }=0.96\text{N m}$$

$\tau =0.96$ N m.

Current sensitivity $=\frac{NAB}{k}$; voltage sensitivity $=\frac{NAB}{kR}$ (same $k$ for both).

(a)

$$\frac{(NAB{)}_2}{(NAB{)}_1}=\frac{42\times 1.8\times 10^{-3}\times 0.50}{30\times 3.6\times 10^{-3}\times 0.25}=\frac{0.0378}{0.027}=1.4$$

(b)

$$\frac{(NAB/R{)}_2}{(NAB/R{)}_1}=1.4\times \frac{R_1}{R_2}=1.4\times \frac{10}{14}=1$$

(a) 1.4. (b) 1 (equal voltage sensitivities).

Why a circle: the magnetic force $\vec{F}=q\vec{v}\times \vec{B}$ is always perpendicular to the velocity. It changes only the direction of motion, never the speed, and its size $evB$ stays constant. A force of constant size that is always perpendicular to the velocity acts as a centripetal force, so the electron moves in a circle.

Radius: $evB=\frac{m{v}^2}{r}$ gives

$$r=\frac{mv}{eB}=\frac{9.1\times 10^{-31}\times 4.8\times 10^6}{1.6\times 10^{-19}\times 6.5\times 10^{-4}}\approx 4.2\times 10^{-2}\text{m}$$

(Some printings show $e=1.5\times 10^{-19}$ C; the correct value is $1.6\times 10^{-19}$ C. With 1.5 the radius comes out about 4.5 cm.)

The magnetic force is perpendicular to the velocity and constant in size, so it is a centripetal force; $r\approx 4.2$ cm.

$$\nu =\frac{v}{2\pi r}=\frac{eB}{2\pi m}=\frac{1.6\times 10^{-19}\times 6.5\times 10^{-4}}{2\pi \times 9.1\times 10^{-31}}\approx 1.8\times 10^7\text{Hz}$$

No, it does not depend on the speed. Since $r=\frac{mv}{eB}$, a faster electron moves in a proportionally larger circle, so the time for one revolution, $T=\frac{2\pi r}{v}=\frac{2\pi m}{eB}$, stays the same.

About 18 MHz; independent of speed, because the radius grows in proportion to the speed.

(a) $A=\pi r^2=\pi (0.08{)}^2=2.01\times 10^{-2}$ m².

$$\tau =NIAB\sin \theta =30\times 6.0\times 2.01\times 10^{-2}\times 1.0\times \sin 60^{\circ }\approx 3.1\text{N m}$$

A counter torque of 3.1 N m is needed.

(b) No. The torque depends only on the area of a planar coil, not on its shape.

(a) About 3.1 N m. (b) No, the torque depends only on the area.