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Moving Charges and Magnetism Class 12th

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?

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Payal Gupta

Contributor-Level 10

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

Since the particle remains un-deflected by electric and magnetic field, we can infer that the electric field is balancing the magnetic field.

Hence eE = evB or v = $\frac{E}{B}$ ………(2)

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

$\frac{e}{m} = \frac{1}{2} \frac{({\frac{E}{B})}^{2}}{V}$ = $\frac{E^{2}}{2 V B^{2}}$ = $\frac{{(9.0 * 10^{5})}^{2}}{2 * 15000 * {0.75}^{2}}$ = 48 $* 10^{6}$ C/kg

This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers ar

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

Since the particle remains un-deflected by electric and magnetic field, we can infer that the electric field is balancing the magnetic field.

Hence eE = evB or v = $\frac{E}{B}$ ………(2)

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

$\frac{e}{m} = \frac{1}{2} \frac{({\frac{E}{B})}^{2}}{V}$ = $\frac{E^{2}}{2 V B^{2}}$ = $\frac{{(9.0 * 10^{5})}^{2}}{2 * 15000 * {0.75}^{2}}$ = 48 $* 10^{6}$ C/kg

This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are ${H e}^{+ +}$ , ${L i}^{+ +}$ etc.

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Moving Charges and Magnetism Class 12th

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.

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Payal Gupta

Contributor-Level 10

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 $* 10^{- 31} k g$

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 = $\sqrt{\frac{2 e V}{m}}$ …….(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 = $\frac{m v}{B e}$ ………………(2)

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

r = $\frac{m}{B e} {[\frac{2 e V}{m}]}^{\frac{1}{2}}$

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

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 t

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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 $* 10^{- 31} k g$

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 = $\sqrt{\frac{2 e V}{m}}$ …….(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 = $\frac{m v}{B e}$ ………………(2)

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

r = $\frac{m}{B e} {[\frac{2 e V}{m}]}^{\frac{1}{2}}$

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

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 $θ$ of 30 $°$ wit initial velocity, the initial velocity will be,

$v_{1}$ = v $\sin ? θ$

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

$r_{1} = \frac{m v_{1}}{B e} = \frac{m v \sin ? θ}{B e} =$ $\frac{9.1 * 10^{- 31}}{0.15 * 1.6 * 10^{- 19}} *$ ${[\frac{2 * 1.6 * 10^{- 19} * 2 * 10^{3}}{9.1 * 10^{- 31}}]}^{\frac{1}{2}} * \sin ? 30 °$

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

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Moving Charges and Magnetism Class 12th

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)

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Payal Gupta

Contributor-Level 10

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 change the direction of velocity, not its magnitude.

(c) 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.

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Moving Charges and Magnetism Class 12th

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.

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Payal Gupta

Contributor-Level 10

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 = $\frac{μ_{0} N I}{l}$ . where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

L = length of the toroid = 2 $π (\frac{r_{1} + r_{2}}{2}$ ) = $π$ (0.25 + 0.26)= 1.6022

B = $(\frac{4 π * 10^{- 7} * 3500 * 11}{1.6022})$ = 3.0 $* 10^{- 2}$ T

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

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Moving Charges and Magnetism Class 12th

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 = $\frac{μ_{0} I R^{2} N}{2 ({x^{2} + R^{2})}^{\frac{3}{2}}}$

(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 $\frac{μ_{0} N I}{R}$ , approximately.

[Such an arrangement to produce a

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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 = $\frac{μ_{0} I R^{2} N}{2 ({x^{2} + R^{2})}^{\frac{3}{2}}}$

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

B =0.72 $\frac{μ_{0} N I}{R}$ , approximately.

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

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Payal Gupta

Contributor-Level 10

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 = $\frac{μ_{0} I R^{2} N}{2 ({x^{2} + R^{2})}^{\frac{3}{2}}}$

where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

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

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

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 $\frac{μ_{0} I R^{2} N}{2} [{{(\frac{R}{2} - d)}^{2} + R^{2}}^{\frac{3}{2}} + {{(\frac{R}{2} + d)}^{2} + R^{2}}^{\frac{3}{2}}]$

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 po

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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 = $\frac{μ_{0} I R^{2} N}{2 ({x^{2} + R^{2})}^{\frac{3}{2}}}$

where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

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

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

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 $\frac{μ_{0} I R^{2} N}{2} [{{(\frac{R}{2} - d)}^{2} + R^{2}}^{\frac{3}{2}} + {{(\frac{R}{2} + d)}^{2} + R^{2}}^{\frac{3}{2}}]$

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:

$B_{1}$ = $\frac{μ_{0} I R^{2} N}{2 ({{\frac{R}{2} + d)}^{2} + R^{2})}^{\frac{3}{2}}}$

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

Magnetic field due to this coil is given as

$B_{2}$ = $\frac{μ_{0} I R^{2} N}{2 ({{\frac{R}{2} + d)}^{2} + R^{2})}^{\frac{3}{2}}}$

Total magnetic field, B = $B_{1}$ + $B_{2}$

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

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

= $\frac{μ_{0} I R^{2} N}{2} [{\frac{5 R^{2}}{4} + d^{2} - R d}^{\frac{3}{2}} + {\frac{5 R^{2}}{4} + d^{2} + R d}^{\frac{3}{2}}]$

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

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

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

= $\frac{μ_{0} I R^{2} N}{2 R^{3}} * {(\frac{4}{5})}^{\frac{3}{2}} [1 - \frac{6 d}{5 R} + 1 + \frac{6 d}{5 R}]$

= ${(\frac{4}{5})}^{\frac{3}{2}} \frac{μ_{0} I N}{R}$

B = 0.72 $\frac{μ_{0} I N}{R}$

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

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Moving Charges and Magnetism Class 12th

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.

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Payal Gupta

Contributor-Level 10

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

$μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

Magnetic field is given by the relation,

B = $μ_{0} n I$ or nI = $\frac{B}{μ_{0}}$ = $\frac{100 * 10^{- 4}}{4 π * 10^{- 7}}$ = 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.

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Moving Charges and Magnetism Class 12th

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.

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Payal Gupta

Contributor-Level 10

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:

$B_{1}$ = $\frac{μ_{0} n_{1} I_{1}}{2 r_{1}}$ , where = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$
$B_{1} = \frac{4 π * 10^{- 7} * 20 * 16}{2 * 0.16}$ = 1.257 $* 10^{- 3}$ T (towards East)

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

$B_{2}$ = $\frac{μ_{0} n_{2} I_{2}}{2 r_{2}}$ , where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

$B_{2} = \frac{4 π * 10^{- 7} * 25 * 18}{2 * 0.10}$ = 2.827 $* 10^{- 3}$ T (towards West)

The net magnetic field B = $B_{2}$ - $B_{1}$ = 2.827 $* 10^{- 3}$ - 1.257 $* 10^{- 3}$ = 1.57 $* 10^{- 3}$ T ( towards West)

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Moving Charges and Magnetism Class 12th

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.)

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Payal Gupta

Contributor-Level 10

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 = $π r^{2}$ = $π * {0.08}^{2}$ = 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, $θ$ = 60°

The coil experiences a toque in the magnetic field, hence it turns.

The counter torque is given by the relation,

$τ = n I B A \sin ? θ$ = 30 $* 6 * 1 * 0.0201 \sin ? 60 °$ = 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.

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Moving Charges and Magnetism Class 12th

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.

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Payal Gupta

Contributor-Level 10

4.12 Let the frequency of revolution = $ν$

Angular frequency = $ω = 2 π ν$

Now, velocity of electron, v = r $ω$

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

evB $= \frac{m v^{2}}{r}$ or eB = $\frac{m v}{r}$ = $\frac{m (r ω)}{r}$ = $\frac{2 π ν m r}{r}$

This frequency is independent of the speed of electron.

$ν = \frac{1.6 * 10^{- 19} * 6.5 * 10^{- 4}}{2 π * 9.1 * 10^{- 31}} = 18.19 * 10^{6}$ Hz = 18.19 MHz

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Moving Charges and Magnetism Class 12th

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^–1normal 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)

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Payal Gupta

Contributor-Level 10

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

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

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

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

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

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

F = evB $\sin ? θ$

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 $\sin ? θ = \frac{m v^{2}}{r}$

r = $\frac{m v}{e B \sin ? θ}$ = $\frac{9.1 * 10^{- 31} * 4.8 * 10^{6}}{1.6 * 10^{- 19} * 6.5 * 10^{- 4} * s i n 90 °}$ = 4.20 $* 10^{- 2}$ m = 4.20 cm

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