Chapter 4 Moving Charges And Magnetism Solutions
Question - 1 : - A circular coil of wireconsisting 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?
Answer - 1 : -
Number of turns on the circularcoil, n = 100
Radius of each turn, r =8.0 cm = 0.08 m
Current flowing in thecoil, I = 0.4 A
Magnitude of the magnetic fieldat the centre of the coil is given by the relation,
Where,
= Permeability of freespace
= 4π × 10–7 T m A–1
Hence, the magnitude of themagnetic field is 3.14 × 10–4 T.
Question - 2 : - A long straight wire carries acurrent of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Answer - 2 : -
Current in the wire, I =35 A
Distance of a point from thewire, r = 20 cm = 0.2 m
Magnitude of the magnetic fieldat this point is given as:
B
Where,
= Permeability of freespace = 4π × 10–7 T m A–1
Hence, the magnitude of themagnetic field at a point 20 cm from the wire is 3.5 × 10–5 T.
Question - 3 : - A long straight wire in thehorizontal plane carries a current of 50 A in north to south direction. Givethe magnitude and direction of B ata point 2.5 m east of the wire.
Answer - 3 : -
Current in the wire, I =50 A
A point is 2.5 m away from theEast of the wire.
Magnitude of the distanceof the point from the wire, r = 2.5 m.
Magnitudeof the magnetic field at that point is given by the relation, B
Where,
= Permeabilityof free space = 4π × 10–7 T m A–1
The point is located normal tothe wire length at a distance of 2.5 m. The direction of the current in thewire is vertically downward. Hence, according to the Maxwell’s right hand thumbrule, the direction of the magnetic field at the given point is verticallyupward.
A horizontal overhead power linecarries a current of 90 A in east to west direction. What is the magnitude anddirection of the magnetic field due to the current 1.5 m below the line?
Answer - 4 : -
Current in the power line, I =90 A
Point is located below the powerline at distance, r = 1.5 m
Hence, magnetic field at thatpoint is given by the relation,
Where,
= Permeabilityof free space = 4π × 10–7 T m A–1
The current is flowing from Eastto West. The point is below the power line. Hence, according to Maxwell’s righthand thumb rule, the direction of the magnetic field is towards the South.
Question - 5 : - What is the magnitude of magneticforce per unit length on a wire carrying a current of 8 A and making an angleof 30º with the direction of a uniform magnetic field of 0.15 T?
Answer - 5 : -
Current in the wire, I =8 A
Magnitude of the uniform magneticfield, B = 0.15 T
Angle between the wire andmagnetic field, θ = 30°.
Magnetic force per unit length onthe wire is given as:
f = BI sinθ
= 0.15 × 8 ×1 × sin30°
= 0.6 N m–1
Hence, the magnetic force perunit length on the wire is 0.6 N m–1.
Question - 6 : - A 3.0 cm wire carrying a currentof 10 A is placed inside a solenoid perpendicular to its axis. The magneticfield inside the solenoid is given to be 0.27 T. What is the magnetic force onthe wire?
Answer - 6 : -
Length of the wire, l =3 cm = 0.03 m
Current flowing in thewire, I = 10 A
Magnetic field, B =0.27 T
Angle between the current andmagnetic field, θ = 90°
Magnetic force exerted on thewire is given as:
F = BIlsinθ
= 0.27 × 10 × 0.03 sin90°
= 8.1 × 10–2 N
Hence, the magnetic force on thewire is 8.1 × 10–2 N. The direction of the force can beobtained from Fleming’s left hand rule.
Question - 7 : - Two long and parallel straightwires A and B carrying currents of 8.0 A and 5.0 A in the same direction areseparated by a distance of 4.0 cm. Estimate the force on a 10 cm section ofwire A.
Answer - 7 : -
Current flowing in wire A, IA =8.0 A
Current flowing in wire B, IB =5.0 A
Distance between the twowires, r = 4.0 cm = 0.04 m
Length of a section of wire A, l =10 cm = 0.1 m
Forceexerted on length l due to the magnetic field is given as:Where,
= Permeability of freespace = 4π × 10–7 T m A–1
The magnitude of force is 2 × 10–5 N.This is an attractive force normal to A towards B because the direction of thecurrents in the wires is the same.
A closely wound solenoid 80 cmlong has 5 layers of windings of 400 turns each. The diameter of the solenoidis 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near itscentre.
Answer - 8 : -
Length of the solenoid, l =80 cm = 0.8 m
There are five layers of windingsof 400 turns each on the solenoid.
Total number of turns on thesolenoid, N = 5 × 400 = 2000
Diameter of the solenoid, D =1.8 cm = 0.018 m
Current carried by thesolenoid, I = 8.0 A
Magnitudeof the magnetic field inside the solenoid near its centre is given by therelation,Where,
= Permeability of freespace = 4π × 10–7 T m A–1
Hence, the magnitude of themagnetic field inside the solenoid near its centre is 2.512 × 10–2 T.
A square coil of side 10 cmconsists of 20 turns and carries a current of 12 A. The coil is suspendedvertically and the normal to the plane of the coil makes an angle of 30º withthe direction of a uniform horizontal magnetic field of magnitude 0.80 T. Whatis the magnitude of torque experienced by the coil?
Answer - 9 : -
Length of a side of the squarecoil, l = 10 cm = 0.1 m
Current flowing in thecoil, I = 12 A
Number of turns on thecoil, n = 20
Angle made by the plane of thecoil with magnetic field, θ = 30°
Strength of magnetic field, B =0.80 T
Magnitude of the magnetic torqueexperienced by the coil in the magnetic field is given by the relation,
τ = n BIA sinθ
Where,
A =Area of the square coil
l × l = 0.1 ×0.1 = 0.01 m2
∴ τ = 20 × 0.8 × 12 × 0.01 × sin30°
= 0.96 N m
Hence, the magnitude of thetorque experienced by the coil is 0.96 N m.
Question - 10 : - Two moving coil meters, M1 andM2 have the following particulars:
R1 =10 Ω, N1 = 30, A1 =3.6 × 10–3 m2, B1 = 0.25T
R2 =14 Ω, N2 = 42, A2 =1.8 × 10–3 m2, B2 = 0.50T
(The spring constants areidentical for the two meters).
Determine the ratio of (a)current sensitivity and (b) voltage sensitivity of M2 and M1.
Answer - 10 : -
For moving coil meter M1:
Resistance, R1 =10 Ω
Number of turns, N1 =30
Area of cross-section, A1 =3.6 × 10–3 m2
Magnetic field strength, B1 =0.25 T
Spring constant K1 = K
For moving coil meter M2:
Resistance, R2 =14 Ω
Number of turns, N2 =42
Area of cross-section, A2 =1.8 × 10–3 m2
Magnetic field strength, B2 =0.50 T
Spring constant, K2 = K
(a) Current sensitivityof M1 is given as:And, current sensitivity of M2 isgiven as:
Ratio 
Hence, the ratio of currentsensitivity of M2 to M1 is 1.4.
(b) Voltagesensitivity for M2 is given as:
And, voltage sensitivity for M1 isgiven as:
