Chapter 4 Moving Charges And Magnetism Solutions
Question - 21 : - The wires which connect thebattery of an automobile to its starting motor carry a current of 300 A (for ashort time). What is the force per unit length between the wires if they are 70cm long and 1.5 cm apart? Is the force attractive or repulsive?
Answer - 21 : -
Current in both wires, I =300 A
Distance between the wires, r =1.5 cm = 0.015 m
Length of the two wires, l =70 cm = 0.7 m
Force between the two wires isgiven by the relation,
Where,
= Permeability of free space = 
Since the direction of thecurrent in the wires is opposite, a repulsive force exists between them.
Question - 22 : - A uniform magnetic field of 1.5 Texists in a cylindrical region of radius10.0 cm, its direction parallel to theaxis along east to west. A wire carrying current of 7.0 A in the north to southdirection passes through this region. What is the magnitude and direction ofthe force on the wire if,
(a) thewire intersects the axis,
(b) thewire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-Sdirection is lowered from the axis by a distance of 6.0 cm?
Answer - 22 : -
Magnetic field strength, B =1.5 T
Radius of the cylindricalregion, r = 10 cm = 0.1 m
Current in the wire passingthrough the cylindrical region, I = 7 A
(a) If thewire intersects the axis, then the length of the wire is the diameter of thecylindrical region.
Thus, l = 2r =0.2 m
Angle between magnetic field andcurrent, θ = 90°
Magnetic force acting on the wireis given by the relation,
F = BIl sin θ
= 1.5 × 7 × 0.2 × sin 90°
= 2.1 N
Hence, a force of 2.1 N acts onthe wire in a vertically downward direction.
(b) Newlength of the wire after turning it to the Northeast-Northwest direction can begiven as: :
Angle between magnetic field andcurrent, θ = 45°
Force on the wire,
F = BIl1 sin θ
Hence, a force of 2.1 N actsvertically downward on the wire. This is independent of angleθbecause l sinθ isfixed.
(c) The wireis lowered from the axis by distance, d = 6.0 cm
Supposewire is passing perpendicularly to the axis of cylindrical magnetic field thenlowering 6 cm means displacing the wire 6 cm from its initial position towardsto end of cross sectional area.
Thus the length of wire in magneticfield will be 16 cm as AB= L =2x =16cm
Now the force,
F = iLB sin90° as the wire will beperpendicular to the magnetic field.
F= 7 × 0.16 × 1.5 =1.68 N
The direction will be given by righthand curl rule or screw rule i.e. vertically downwards.
Question - 23 : - A uniform magnetic field of 3000G is established along the positive z-direction. A rectangular loopof sides 10 cm and 5 cm carries a current of 12 A. What is the torque on theloop in the different cases shown in Fig. 4.28? What is the force on each case?Which case corresponds to stable equilibrium?
Answer - 23 : -
Magnetic field strength, B =3000 G = 3000 × 10−4 T = 0.3 T
Length of the rectangular loop, l =10 cm
Width of the rectangularloop, b = 5 cm
Area of the loop,
A = l × b =10 × 5 = 50 cm2 = 50 × 10−4 m2
Current in the loop, I =12 A
Now, taking the anti-clockwisedirection of the current as positive and vise-versa:
(a) Torque, 
Fromthe given figure, it can be observed that A is normal tothe y-z plane and B is directed alongthe z-axis.The torque is 
N m along thenegative y-direction. The force on the loop is zero because theangle between A and B is zero.
(b) Thiscase is similar to case (a). Hence, the answer is the same as (a).
(c) Torque 
From the given figure, it can beobserved that A is normal to the x-z planeand B is directed along the z-axis.
The torque is N m along thenegative x direction and the force is zero.
(d) Magnitude of torqueis given as:
Torque is N m at an angle of 240°with positive x direction. The force is zero.
(e) Torque Hence, the torque is zero. Theforce is also zero.
(f) Torque Hence, the torque is zero. Theforce is also zero.
Incase (e), the direction of 
and 
is the same and the angle betweenthem is zero. If displaced, they come back to an equilibrium. Hence, itsequilibrium is stable.A circular coil of 20 turns andradius 10 cm is placed in a uniform magnetic field of 0.10 T normal to theplane of the coil. If the current in the coil is 5.0 A, what is the
(a) totaltorque on the coil,
(b) totalforce on the coil,
(c) averageforce on each electron in the coil due to the magnetic field?
(The coil is made of copper wireof cross-sectional area 10−5 m2, and the freeelectron density in copper is given to be about 1029 m−3.)
Answer - 24 : -
Number of turns on the circularcoil, n = 20
Radius of the coil, r =10 cm = 0.1 m
Magnetic field strength, B =0.10 T
Current in the coil, I =5.0 A
(a) The totaltorque on the coil is zero because the field is uniform.
(b) The totalforce on the coil is zero because the field is uniform.
(c) Cross-sectionalarea of copper coil, A = 10−5 m2
Number of free electrons percubic meter in copper, N = 1029 /m3
Charge on the electron, e =1.6 × 10−19 C
Magnetic force, F = Bevd
Where, vd = Driftvelocity of electrons
Hence, the average force on each electron is 