Chapter 4 Moving Charges And Magnetism Solutions
Question - 11 : - In a chamber, a uniform magneticfield of 6.5 G (1 G = 10–4 T) is maintained. An electron isshot into the field with a speed of 4.8 × 106 m s–1 normalto the field. Explain why the path of the electron is a circle. Determine theradius of the circular orbit. (e = 1.6 × 10–19 C, me=9.1×10–31 kg)
Answer - 11 : -
Magnetic field strength, B =6.5 G = 6.5 × 10–4 T
Speed of the electron, v =4.8 × 106 m/s
Charge on the electron, e =1.6 × 10–19 C
Mass of the electron, me =9.1 × 10–31 kg
Angle between the shot electronand magnetic field, θ = 90°
Magnetic force exerted on theelectron in the magnetic field is given as:
F = evB sinθ
This force provides centripetalforce to the moving electron. Hence, the electron starts moving in a circularpath of radius r.
Hence, centripetal force exerted onthe electron,
In equilibrium, the centripetalforce exerted on the electron is equal to the magnetic force i.e.,
Hence, the radius of the circularorbit of the electron is 4.2 cm.
Question - 12 : - In Exercise 4.11 obtain thefrequency of revolution of the electron in its circular orbit. Does the answerdepend on the speed of the electron? Explain.
Answer - 12 : -
Magnetic field strength, B =6.5 × 10−4 T
Charge of the electron, e =1.6 × 10−19 C
Mass of the electron, me =9.1 × 10−31 kg
Velocity of the electron, v =4.8 × 106 m/s
Radius of the orbit, r =4.2 cm = 0.042 m
Frequency of revolution of theelectron = ν
Angular frequency of the electron= ω = 2πν
Velocity of the electron isrelated to the angular frequency as:
v = rω
In the circular orbit, themagnetic force on the electron is balanced by the centripetal force. Hence, wecan write:
This expression for frequency isindependent of the speed of the electron.
Onsubstituting the known values in this expression, we get the frequency a
Hence, the frequency of theelectron is around 18 MHz and is independent of the speed of the electron.
Question - 13 : - (a) Acircular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A issuspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T.The field lines make an angle of 60º with the normal of the coil. Calculate themagnitude of the counter torque that must be applied to prevent the coil fromturning.
(b) Wouldyour answer change, if the circular coil in (a) were replaced by a planar coilof some irregular shape that encloses the same area? (All other particulars arealso unaltered.)
Answer - 13 : -
(a) Numberof turns on the circular coil, n = 30
Radius of the coil, r =8.0 cm = 0.08 m
Area of the coil 
Current flowing in thecoil, I = 6.0 A
Magnetic field strength, B =1 T
Angle between the field lines andnormal with the coil surface,
θ =60°
The coil experiences a torque inthe magnetic field. Hence, it turns. The counter torque applied to prevent thecoil from turning is given by the relation,
τ = n IBA sinθ …(i)
= 30 × 6 × 1 × 0.0201 × sin60°
= 3.133 N m
(b) Itcan be inferred from relation (i) that the magnitude of the appliedtorque is not dependent on the shape of the coil. It depends on the area of thecoil. Hence, the answer would not change if the circular coil in the above caseis replaced by a planar coil of some irregular shape that encloses the samearea.
Question - 14 : - Two concentric circular coils Xand Y of radii 16 cm and 10 cm, respectively, lie in the same vertical planecontaining the north to south direction. Coil X has 20 turns and carries acurrent of 16 A; coil Y has 25 turns and carries a current of 18 A. The senseof the current in X is anticlockwise, and clockwise in Y, for an observerlooking at the coils facing west. Give the magnitude and direction of the netmagnetic field due to the coils at their centre.
Answer - 14 : -
Radius of coil X, r1 =16 cm = 0.16 m
Radius of coil Y, r2 =10 cm = 0.1 m
Number of turns of on coilX, n1 = 20
Number of turns of on coilY, n2 = 25
Current in coil X, I1 =16 A
Current in coil Y, I2 =18 A
Magnetic field due to coil X attheir centre is given by the relation,
Where,
= Permeability of freespace = 
Magnetic field due to coil Y attheir centre is given by the relation,
Hence, net magnetic field can beobtained as:
Question - 15 : - A magnetic field of 100 G (1 G =10−4 T) is required which is uniform in a region of lineardimension 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 thenumber of turns per unit length that can be wound round a core is at most 1000turns m−1. Suggest some appropriate design particulars of a solenoidfor the required purpose. Assume the core is not ferromagnetic
Answer - 15 : -
Magnetic field strength, B =100 G = 100 × 10−4 T
Number of turns per unitlength, n = 1000 turns m−1
Current flowing in thecoil, I = 15 A
Permeability of free space, =
Magneticfield is given by the relation,If the length of the coil istaken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then thesevalues are not unique for the given purpose. There is always a possibility ofsome adjustments with limits.
For a circular coil ofradius R and N turns carrying current I,the magnitude of the magnetic field at a point on its axis at a distance x fromits centre is given by,
(a) Showthat this reduces to the familiar result for field at the centre of the coil.
(b) Considertwo parallel co-axial circular coils of equal radius R, and numberof turns N, carrying equal currents in the same direction, and separatedby a distance R. Show that the field on the axis around themid-point between the coils is uniform over a distance that is small ascompared to R, and is given by,
, approximately.
[Such an arrangement to produce anearly uniform magnetic field over a small region is known as Helmholtzcoils.]
Answer - 16 : -
Radius of circular coil = R
Number of turns on the coil= N
Current in the coil = I
Magnetic field at a point on itsaxis at distance x is given by the relation,
Where,
= Permeability of free space
(a) If themagnetic field at the centre of the coil is considered, then x =0.
This is the familiar result formagnetic field at the centre of the coil.
(b) Radii oftwo parallel co-axial circular coils = R
Number of turns on each coil= N
Current in both coils = I
Distance between both the coils= R
Let us consider point Q atdistance d from the centre.
Then,one coil is at a distance of 
from point Q.
agnetic field at point Q isgiven as:
Also, the other coil is at a distance of 
from point Q.
Magnetic field due to this coilis given as:
Total magnetic field,
Hence, it is proved that thefield on the axis around the mid-point between the coils is uniform.
Question - 17 : - A toroid has a core(non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which3500 turns of a wire are wound. If the current in the wire is 11 A, what is themagnetic field (a) outside the toroid, (b) inside the core of the toroid, and(c) in the empty space surrounded by the toroid.
Answer - 17 : -
Inner radius of the toroid, r1 =25 cm = 0.25 m
Outer radius of the toroid, r2 =26 cm = 0.26 m
Number of turns on the coil, N =3500
Current in the coil, I =11 A
(a) Magneticfield outside a toroid is zero. It is non-zero only inside the core of a toroid.
(b) Magneticfield inside the core of a toroid is given by the relation,
B = 
Where,
= Permeability of free space =
l =length of toroid
(c) Magneticfield in the empty space surrounded by the toroid is zero.
Question - 18 : - An electron emitted by a heatedcathode and accelerated through a potential difference of 2.0 kV, enters aregion with uniform magnetic field of 0.15 T. Determine the trajectory of theelectron if the field (a) is transverse to its initial velocity, (b) makes anangle of 30º with the initial velocity.
Answer - 18 : -
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 × 103 V
Thus, kinetic energy of theelectron = eV
Where,
v =velocity of the electron
(a) Magneticforce on the electron provides the required centripetal force of the electron.Hence, the electron traces a circular path of radius r.
Magnetic force on the electron isgiven by the relation,
B ev
Centripetalforce 
From equations (1) and (2), weget
Hence, the electron has acircular trajectory of radius 1.0 mm normal to the magnetic field.
(b) Whenthe field makes an angle θ of 30° with initial velocity, theinitial velocity will be,
From equation (2), we can writethe expression for new radius as:
Hence, the electron has a helicaltrajectory of radius 0.5 mm along the magnetic field direction.
Question - 19 : - A magnetic field set up usingHelmholtz coils (described in Exercise 4.16) is uniform in a small region andhas a magnitude of 0.75 T. In the same region, a uniform electrostatic field ismaintained in a direction normal to the common axis of the coils. A narrow beamof (single species) charged particles all accelerated through 15 kV enters thisregion in a direction perpendicular to both the axis of the coils and theelectrostatic field. If the beam remains undeflected when the electrostaticfield is 9.0 × 105 V m−1, make a simple guess as towhat the beam contains. Why is the answer not unique?
Answer - 19 : -
Magnetic field, B =0.75 T
Accelerating voltage, V =15 kV = 15 × 103 V
Electrostatic field, E =9 × 105 V m−1
Mass of the electron = m
Charge of the electron = e
Velocity of the electron = v
Kinetic energy of the electron= eV
Since the particle remainsundeflected by electric and magnetic fields, we can infer that the forceon the charged particle due to electric field is balancing the force onthe charged particle due to magnetic field.
Putting equation (2) in equation(1), we get
This value of specificcharge e/m is equal to the value of deuteron ordeuterium ions. This is not a unique answer. Other possible answers are He++, Li++,etc.
Question - 20 : - A straight horizontal conductingrod of length 0.45 m and mass 60 g is suspended by two vertical wires at itsends. A current of 5.0 A is set up in the rod through the wires.
(a) Whatmagnetic field should be set up normal to the conductor in order that thetension in the wires is zero?
(b) Whatwill be the total tension in the wires if the direction of current is reversedkeeping the magnetic field same as before? (Ignore the mass of the wires.) g =9.8 m s−2.
Answer - 20 : -
Length of the rod, l =0.45 m
Mass suspended by thewires, m = 60 g = 60 × 10−3 kg
Acceleration due to gravity, g =9.8 m/s2
Current in the rod flowingthrough the wire, I = 5 A
(a) Magnetic field (B)is equal and opposite to the weight of the wire i.e.,
A horizontal magnetic field of0.26 T normal to the length of the conductor should be set up in order to getzero tension in the wire. The magnetic field should be such that Fleming’s lefthand rule gives an upward magnetic force.
(b) If the directionof the current is revered, then the force due to magnetic field and the weightof the wire acts in a vertically downward direction.
∴Total tension in the wire = BIl + mg
