Class 12th

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $*{10}^{-3}$ kg $*{10}^{-3}$

Hence, the current flowing through the solenoid is 108 A.

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.