Physics

4.24 Magnetic field strength, B = 3000G = 3000 $*{10}^{-4}$ T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop, A = l $*b$ = 10 $*5$ = 50 ${cm}^2$ = 50 $*{10}^{-4}$ $m^2$

Current in the loop, I = 12 A

Assume that the anti-clockwise direction of the current is positive and vice versa.

Torque, $\stackrel{?}{\tau }$ = $\stackrel{?}{A}$ $*\stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along z-axis.

$\tau$ = 12*( 50 $*{10}^{-4})\stackrel{?}{i}*0.3\stackrel{?}{k}$ = $-1.8*{10}^{-2}\stackrel{?}{j}$ Nm

The Torque $1.8*{10}^{-2}$ is Nm along the negative y-direction.

The force on the loop is zero because the angle between A & B is zero.

This case is similar to case (a). The answer is same as case (a)

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $*\stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the x-z plane and B is directed along z-axis.

= - 12 $*$ (50 $*{10}^{-4})\stackrel{?}{i}*0.3\stackrel{?}{k}$ = $1.8*{10}^{-2}$ Nm

The Torque is $1.8*{10}^{-2}$ Nm at an angle 240 $°$ with positive x direction. The force is zero

Magnitude of the torque is given as:

| $\left|\tau \right|$ |= IAB|

12 $*$ 50 $*{10}^{-4}*0.3$ = $1.8*{10}^{-2}$ Nm

The Torque is $1.8*{10}^{-2}$ is Nm at an angle 240 with positive x direction. The force is zero.

Torque, $\left|\tau \right|$ = I $\stackrel{?}{A}$ $*\stackrel{?}{B}$

= (12 $*$ 50 $*{10}^{-4})\stackrel{?}{k}$ $*0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $*\stackrel{?}{B}$

= (12 $*$ 50 $*{10}^{-4})\stackrel{?}{k}$ $*0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

In the case of (e), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is the same and the angle between them is zero. If displaced, they come back to equilibrium. Hence, its equilibrium is stable.

In the case of (f), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is opposite. The angle between them is 180 $°$ . If disturbed, it does not come back to its original position, hence the equilibrium is unstable.

4.20 Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 $*{10}^{-3}$ V

Electrostatic field, E = 9.0 $*{10}^5$ V/m

Let the mass of electron = m, Charge of the electron = e, Velocity of the electron = v

Then kinetic energy of the electron = eV

$\frac{1}{2}$ m $v^2$ = eV or $\frac{e}{m}$ = $\frac{v^2}{2V}$ ………….(1)

4.19 Magnetic field strength, B = 0.15 T

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

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

Potential difference, V = 2.0 kV = 2 $*{10}^3$ 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 centripetal force of the electron. Hence, electron traces a circular path of radius r

Magnetic force on the electron = Bev

Centripetal force $\frac{m{v}^2}{r}$

Hence, Bev = $\frac{m{v}^2}{r}$

r = $\frac{mv}{Be}$ ………………(2)

From equation (1) and (2), we get

r = $\frac{m}{Be}{\left[\frac{2eV}{m}\right]}^{\frac{1}{2}}$

= $\frac{9.1\mathrm{}*{10}^{-31}}{0.15*1.6\mathrm{}*{10}^{-19}}*$ ${\left[\frac{2*1.6\mathrm{}*{10}^{-19}*2\mathrm{}*{10}^3}{9.1\mathrm{}*{10}^{-31}}\right]}^{\frac{1}{2}}$

r = 1.006 $*{10}^{-3}$ m = 1.0 mm

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

When the field makes an angle $\theta$ of 30 $°$ wit initial velocity, the initial velocity will be,

$v_1$ = v $\sin ?\theta$

$\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{m}$ equation (2), we can write the expression for the new radius as :

$r_1=\frac{m{v}_1}{Be}=\frac{mv\sin ?\theta }{Be}=$ $\frac{9.1\mathrm{}*{10}^{-31}}{0.15*1.6\mathrm{}*{10}^{-19}}*$ ${\left[\frac{2*1.6\mathrm{}*{10}^{-19}*2\mathrm{}*{10}^3}{9.1\mathrm{}*{10}^{-31}}\right]}^{\frac{1}{2}}*\sin ?30°$

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.

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 $\pi *{10}^{-7}$ 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 is the familiar result for magnetic field at the centre of the coil.

Radius of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both the coils = I $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

Distance between both the coils = R

Let us consider point Q at a distance d from the centre.

Then one coil is at a distance of + d from point Q

Magnetic field at point Q is given as:

$B_1$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

$\mathrm{}\mathrm{A}\mathrm{l}\mathrm{s}\mathrm{o}$ , the other coil is at a distance of $\frac{R}{2}$ – d from point Q,

Magnetic field due to this coil is given as

$B_2$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

Total magnetic field, B = $B_1$ + $B_2$

= $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$ + $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}-d)}^2+R^2)}^{\frac{3}{2}}}$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{\{\frac{5R^2}{4}+d^2-Rd\}}^{\frac{3}{2}}+{\{\frac{5R^2}{4}+d^2+Rd\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}*{\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1+\frac{4}{5}\frac{d^2}{R^2}-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d^2}{R^2}+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$

For d $?R$ , neglecting the factor , we get

= $\frac{{\mu }_{0}I{R}^{2}N}{2}*{\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$