Moving Charges and Magnetism

4.19 Magnetic field strength, B = 0.15 T

Charge on the electron, e = 1.6 $\times {10}^{-19}$ C

Mass of the electron, m = 9.1 $\times {10}^{-31}kg$

Potential difference, V = 2.0 kV = 2 $\times {10}^3$ V

Thus the kinetic energy of the electron = eV = $\frac{1}{2}$ m $v^2$ , where v = velocity of electron

v = $\sqrt{\frac{2eV}{m}}$ …….(1)

Magnetic force on the electron provides the required centr

4.18 (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.

(b) Yes, the final speed of the particle will be equal to its initial speed. This because magnetic force can

4.17 Inner radius of the toroid, $r_1$ = 25 cm = 0.25 m

Outer radius of the toroid, $r_2$ = 26 cm = 0.26 m

Number of turns on the coil, N = 3500

Current in the coil, I = 11 A

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

Magnetic field inside the core of a toroid is gi

4.16 Radius of the circular coil = R

Number of turns on the coil = N

Current in the coil= I

Magnetic field at a point on its axis at a distance x is given as:

B = $\frac{{\mathit{\mu }}_0\mathit{I}{\mathit{R}}^2\mathit{N}}{2({{\mathit{x}}^2+\mathit{}{\mathit{R}}^2)}^{\frac{3}{2}}}$

where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

If the magnetic field at the centre of the coil is considered, then x = 0, then

B = $\frac{{\mathit{\mu }}_0\mathit{I}{\mathit{R}}^2\mathit{N}}{2({0+\mathit{}{\mathit{R}}^2)}^{\frac{3}{2}}}$ = $\frac{{\mathit{\mu }}_0\mathit{I}\mathit{N}}{2\mathit{R}}$

This

4.15 Magnetic field strength, B = 100 G = 100 ${10}^{-4}$ T

Number of turns per unit length, n = 1000 turns / m

Current flowing in the coil, I = 15 A

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Magnetic field is given by the relation,

B = ${\mu }_{0}nI$ or nI = $\frac{B}{{\mu }_0}$ = $\frac{100\times \mathrm{}{10}^{-4}}{4\pi \times {10}^{-7}}$ = 7957.75 A $m^{-1}\approx 8000$ A $m^{-1}$

If the length of the coil is taken as 50 cm,

4.14 For coil X:

Radius, $r_1$ = 16 cm = 0.16 m

Number of turns, $n_1$ = 20

Current, $I_1$ = 16 A

For coil Y:

Radius, $r_2$ = 10 cm = 0.10 m

Number of turns, $n_2$ = 25

Current, $I_2$ = 18 A

Magnetic field due to coil X at their centre is given by the relation:

$B_1$ = $\frac{{\mu }_0{n}_1{I}_1}{2r_1}$ , where = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$
$B_1=\frac{4\pi \times {10}^{-7}\mathrm{}\times 20\times 16}{2\times 0.16}$ = 1.257 $\times {10}^{-3}$ T (to

4.13 Number of turns of the circular coil, n = 30

Radius of the coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $\pi r^2$ = $\pi \times {0.08}^2$ = 0.0201 $m^2$

Current flowing in the coil, I = 6.0 A

Magnetic field strength, B = 1 T

Angle between the field line and normal of the coil surface, $\theta$ = 60°

The coil experiences a toque in the

4.12 Let the frequency of revolution = $\nu$

Angular frequency = $\omega =2\pi \nu$

Now, velocity of electron, v = r $\omega$

Since in circular orbit, magnetic force is balanced by the centripetal force, we can write

evB $=\frac{m{v}^2}{r}$ or eB = $\frac{mv}{r}$ = $\frac{m\left(\mathrm{r}\omega \right)}{r}$ = $\frac{2\pi \nu mr}{r}$

This frequency is independent of the speed of electron.

$\nu =\frac{1.6\mathrm{}\times {10}^{-19}\times 6.5\times {10}^{-4}}{2\pi \times 9.1\mathrm{}\times {10}^{-31}}=18.19\times {10}^6$ Hz = 18.19 MHz

4.11 Magnetic field strength, B = 6.5 G = 6.5 $\times {10}^{-4}$ T

Speed of electron, v = 4.8 $\times {10}^6$ m/s

Charge of electron, e = 1.6 $\times {10}^{-19}$ C

Mass of electron, m = 9.1 $\times {10}^{-31}$ kg

Angle between the shot electron and the magnetic field, $\theta$ = 90 $°$

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB $\sin ?\theta$

This force p

4.10 For moving coil meter M1

Current sensitivity of $M_1$ is given as $I_{s1}$ = $\frac{N_1{B}_1{A}_1}{K_1}$ and

for $M_2$ is given as $I_{s2}$ = $\frac{N_2{B}_2{A}_2}{K_2}$ where $K_1$ and $K_2$ are spring constants for both $M_1$ & $M_2$ . It is given $K_1$ = $K_2$

The ratio of current sensitivity is given as $\frac{I_{s2}\mathrm{}}{I_{s1}}$ = $\frac{N_2{B}_2{A}_2}{N_1{B}_1{A}_1}$ = $\frac{42\times 0.5\times 1.8\mathrm{}\times \mathrm{}{10}^{-3}}{30\times 0.25\times 3.6\times {10}^{-3}}$ = 1.4

Voltage sensitivity of $M_{1}and{M}_2$ is given by $V_{S1}$ = $\frac{N_1{B}_1{A}_1}{K_1{R}_1}$ and $V_{S2}$ = $\frac{N_2{B}_2{A}_2}{K_2{R}_2}$

The ratio of vo