Class 12 Physics Chapter 4 - Moving Charges and Magnetism NCERT Solutions: Questions and Answers PDF

Q.4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Ans.4.1 Number of turns of the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given as

$\left|\vec{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2\pi nI}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$
Hence, $\left|\vec{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\pi \times 100\times 0.4}{0.08}$ = 3.14 $\times {10}^{-4}$ T

 

Q.4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Ans.4.2 Current in the wire, I = 35 A

Distance of a point from the wire, r = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as

$\left|\vec{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

Hence, $\left|\vec{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 35}{0.2}$ = 3.50 $\times {10}^{-5}$ T

 

Q.4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Ans.4.3 Current in the wire, I = 50 A

The distance of the point from the wire, r = 2.5 m

Magnitude of the magnetic field at this point is given as

$\left|\vec{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

Hence, $\left|\vec{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 50}{2.5}$ = 4.0 $\times {10}^{-6}$ T

The direction of the current in the wire is vertically downward. Hence, according to Maxwell’s right hand rule, the direction of the magnetic field at the given point is vertically upward.

 

Q.4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Ans.4.4 Current in the power line, I = 90 A

Point is located below power line at a distance, r = 1.5 m

Magnitude of the magnetic field at this point is given as

$\left|\vec{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space $\pi \times {10}^{-7}$ = 4 Tm $A^{-1}$

Hence, $\left|\vec{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 90}{1.5}$ = 1.2 $\times {10}^{-5}$ T

 

Q.4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Ans.4.5 Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and the magnetic field, $\theta$ = 30 $°$

Magnetic force per unit length of the wire is given as

f = BI $\sin \theta$ = 0.15 $\times 8\times \sin 30°$ = 0.6 N/m

 

Q.4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Ans.4.6 Length of the wire, l = 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

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

Magnetic force exerted on the wire is given as

F= BI $l\sin \theta$ = 0.27 $\times 10\times 0.03\sin 90°$ = 8.1 $\times {10}^{-2}$ N

 

Q.4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Ans.4.7 Current flowing in wire A, $I_A$ = 8.0 A

Current flowing in wire B, $I_B$ = 5.0 A

Distance between two wires, r = 4.0 cm = 0.04 m

Length of the section of the wire A, L = 10 cm = 0.1 m

Force exerted on length L due to magnetic field is given as:

F = $\frac{{\mu }_0{I}_A{I}_{B}L}{2\pi r}$ , where ${\mu }_0$ = permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

F = $\frac{4\pi \times {10}^{-7}\times 8\times 5\times 0.1}{2\pi \times 0.04}$ = 2 $\times {10}^{-5}$ N

The magnitude of force is 2 $\times {10}^{-5}$ N. This is an attractive force normal to A, towards B. Because the direction of the currents in both the wire is same.

 

Q.4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Ans.4.8 Length of the solenoid, l = 80 cm = 0.8 m

Total number of turns in 5 layers, n = 5 $\times 400$ = 2000

Diameter of the solenoid, D = 1.8 cm = 0.018 m

Current carrying by the solenoid, I = 8.0 A

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

B = $\frac{{\mu }_{0}NI}{l}$ , where ${\mu }_0$ = permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

B = $\frac{4\pi \times {10}^{-7}\times 2000\times 8}{0.8}$ = 2.51 $\times {10}^{-2}$ T

 

Q.4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Ans.4.9 Length of a side of the square coil, l = 10 cm = 0.1 m

Current flowing through the coil, I = 12 A

Number of turns of the coil, n = 20

Angle made by the plane of the coil with magnetic field, $\theta$ = 30 $°$

Strength of the magnetic field, B = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by,

$\tau$ = nBIA $\sin \theta$ , where A = Area of the square coil = 0.1 $\times 0.1=0.01m^2$

 

$\tau =20\times 0.8\times 12\times 0.01\times \sin 30°$ = 0.96 Nm

Q.4.10 Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Ans.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 voltage sensitivity $\frac{V_{S2}}{V_{S1}}$ = $\frac{N_2{B}_2{A}_2{K}_1{R}_1}{N_1{B}_1{A}_1{K}_2{R}_2}$ = $\frac{N_2{B}_2{A}_2{R}_1}{N_1{B}_1{A}_1{R}_2}$

= $\frac{42\times 0.5\times 1.8\mathrm{}\times \mathrm{}{10}^{-3}\times 10}{30\times 0.25\times 3.6\times {10}^{-3}\times 14}$ = 1

 

Q.4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

(e = 1.6 × 10–19 C, me = 9.1×10–31 kg)

Ans.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 provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r

The centripetal force exerted on electron, $F_c$ = $\frac{m{v}^2}{r}$

In equilibrium, the centripetal force exerted on electron = magnetic force on the electron

F = $F_c$

evB $\sin \theta =\frac{m{v}^2}{r}$

r = $\frac{mv}{\mathrm{e}\mathrm{B}\sin \theta }$ = $\frac{9.1\mathrm{}\times {10}^{-31}\times 4.8\mathrm{}\times {10}^6}{1.6\mathrm{}\times {10}^{-19}\times 6.5\times {10}^{-4}\times sin90°}$ = 4.20 $\times {10}^{-2}$ m = 4.20 cm

 

Q.4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Ans.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

Q.4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Ans.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 magnetic field, hence it turns.

The counter torque is given by the relation,

$\tau =nIBA\sin \theta$ = 30 $\times 6\times 1\times 0.0201\sin 60°$ = 3.133 N

Since the magnitude of the torque is not dependent on the shape of the coil, it depends only on the area. Hence he answer will not change if the circular coil is replaced with a planar coil of same area.

 

Q.4.14 Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Ans.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 (towards East)

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

$B_2$ = $\frac{{\mu }_0{n}_2{I}_2}{2r_2}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

$B_2=\frac{4\pi \times {10}^{-7}\mathrm{}\times 25\times 18}{2\times 0.10}$ = 2.827 $\times {10}^{-3}$ T (towards West)

The net magnetic field B = $B_2$ - $B_1$ = 2.827 $\times {10}^{-3}$ - 1.257 $\times {10}^{-3}$ = 1.57 $\times {10}^{-3}$ T ( towards West)

 

Q.4.15 A magnetic field of 100 G (1 G = 10–4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10–3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m–1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Ans.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 , radius 4 cm, number of turns 400 and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

 

Q.4.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

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

(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

B =0.72 $\frac{{\mathit{\mu }}_0\mathit{N}\mathit{I}}{\mathit{R}}$ , approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Ans.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 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 = $B_1$ + $B_2$

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

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

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

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

For d $\ll R$ , neglecting the factor , we get

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

= $\frac{{\mu }_{0}I{R}^{2}N}{2R^3}\times {\left(\frac{4}{5}\right)}^{\frac{3}{2}}\left[1-\frac{6d}{5R}+1+\frac{6d}{5R}\right]$

= ${\left(\frac{4}{5}\right)}^{\frac{3}{2}}\frac{{\mu }_{0}IN}{R}$

B = 0.72 $\frac{{\mu }_{0}IN}{R}$

Hence, it is proved that the field on the axis around mid-point between the coils is uniform.

 

Q.4.17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.

Ans.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 given by the relation,

B = $\frac{{\mu }_{0}NI}{l}$ . where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

L = length of the toroid = 2 $\pi (\frac{r_1+r_2}{2}$ ) = $\pi$ (0.25 + 0.26)= 1.6022

B = $\left(\frac{4\pi \times {10}^{-7}\times 3500\times 11}{1.6022}\right)$ = 3.0 $\times {10}^{-2}$ T

Magnetic field in the empty space surrounded by the toroid is zero.

 

Q.4.18 Answer the following questions:

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Ans.4.18 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.

Yes, the final speed of the particle will be equal to its initial speed. This because magnetic force can change the direction of velocity, not its magnitude.

This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the south. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.

 

Q.4.19 An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field

(a) is transverse to its initial velocity,

(b) makes an angle of 30º with the initial velocity.

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

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

r = 1.006 $\times {10}^{-3}$ 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 :

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

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

 

Q.4.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Ans.4.20 Magnetic field, B = 0.75 T

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

Electrostatic field, E = 9.0 $\times {10}^5$ 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)