This chapter delves into the interplay between electricity and magnetism, exploring the magnetic effects of moving charges and currents.
Key Concepts:
Magnetic Force on a Moving Charge: A charged particle moving in a magnetic field experiences a force perpendicular to both its velocity and the magnetic field direction.
Magnetic Field due to a Current: A current-carrying conductor generates a magnetic field around it, following the right-hand thumb rule.
Force Between Current-Carrying Conductors: Parallel currents attract, while anti-parallel currents repel each other.
Moving Coil Galvanometer: A device that measures electric current by using the magnetic force on a current-carrying coil.
Ampere’s Circuital Law: Relates the circulation of the magnetic field around a closed loop to the total current passing through the loop.
This foundational knowledge underpins numerous technological applications, from electric motors to magnetic resonance imaging.
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 :The magnetic field at the center of a circular coil of radius R, carrying a current I, and having N turns is given by:
B = (μ₀NI) / (2R)
Where:
* μ₀ denote permeability of free space
(4π × 10⁻⁷ T m/A)
* N denote number of turns
(100)
* I denote current
(0.40 A)
* R denote radius of the coil
(8.0 cm = 0.08 m)
Substituting the values:
B = (4π × 10⁻⁷ T m/A × 100 × 0.40 A) / (2 × 0.08 m) ≈ 3.14 × 10⁻⁴ T
Therefore, the magnitude of the magnetic field at the center of the coil is approximately 3.14 × 10⁻⁴ T.
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 :The strength of the magnetic field (B) at a distance (r) from a long, straight wire carrying a current (I) is given by following formula
B = (μ₀I) / (2πr)
Where:
* μ₀ denote permeability of
The permeability of free space is 4π × 10⁻⁷ tesla meter per ampere.
* I denote current
(35 A)
* r denote distance from the wire
(20 cm = 0.2 m)
Substituting the values:
B = (4π × 10⁻⁷ T m/A × 35 A) / (2π × 0.2 m) ≈ 3.5 × 10⁻⁵ T
Therefore, the magnitude of the magnetic field at the point 20 cm from the wire is approximately 3.5 × 10⁻⁵ T.
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 :The magnetic field (B) around a long, straight wire carrying a current (I) has a magnitude given by the formula:
B = (μ₀I) / (2πr)
Where:
* μ₀ denote permeability of
The permeability of free space is 4π × 10⁻⁷ tesla meter per ampere.
* I denote current (50 A)
* r denote distance from the wire (2.5 m)
Substituting the values:
B = (4π × 10⁻⁷ T m/A × 50 A) / (2π × 2.5 m)
= 4 × 10⁻⁶ T
The right-hand thumb rule allows us to visualize and identify the direction of the magnetic field. Point your thumb in the direction of the current (north to south), and your fingers will curl in the direction of the magnetic field around the wire. At a point 2.5 m east of the wire, your fingers will point vertically downwards.
Therefore, the magnitude of the magnetic field at the point is 4 × 10⁻⁶ T, and its direction is vertically downwards.
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 :The magnetic field (B) around a long, straight wire carrying a current (I) has a magnitude given by the formula:
B = (μ₀I) / (2πr)
Where:
* μ₀ denote permeability of
The permeability of free space is 4π × 10⁻⁷ tesla meter per ampere.
* I denote current (90 A)
* r denote distance from the wire (1.5 m)
Substituting the values:
B = (4π × 10⁻⁷ T m/A × 90 A) / (2π × 1.5 m) = 1.2 × 10⁻⁵ T
The right-hand thumb rule allows us to visualize and identify the direction of the magnetic field. Point your thumb in the direction of the current (east to west), and your fingers will curl in the direction of the magnetic field around the wire. At a point 1.5 m below the wire, your fingers will point towards the north.
Therefore, the magnitude of the magnetic field at the point is 1.2 × 10⁻⁵ T**, and its direction is north.
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 :A wire carrying current I in a uniform magnetic field B experiences a force:
F = BIL sinθ
Where:
F denote force
B denote magnetic field strength (0.15 T)
I denote current (8 A)
L denote length of the wire
θ = angle between the current direction and the magnetic field (30°)
Determine force per unit length, we divide the force by the length of the wire:
The force per unit length is given by BI sinθ.
Substituting the given values:
F/L = (0.15 T) × (8 A) × sin(30°) = 0.6 N/m
Therefore, the magnetic force acting on the wire is 0.6 Newtons per meter.
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 :To calculate the force, we use the formula:
F = BIl sinθ
where:
F denote magnetic force
B denote magnetic field strength (0.27 T)
I denote current flowing through the wire (10 A)
l = length of the wire (0.03 m)
θ = angle between the current direction and the magnetic field (90°)
Substituting the values, we get:
F = 0.27 T x 10 A x 0.03 m x sin 90°
F = 8.1 x 10^-2 N
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 :The magnetic force experienced per unit length by two parallel, current-carrying conductors is given by:
F/L = (μ₀I₁I₂) / (2πd)
Where:
* μ₀ = permeability of free space
(4π × 10⁻⁷ T m/A)
*The symbols I₁ and I₂ are used to designate the currents in the two wires.
(8.0 A and 5.0 A)
* d = distance between the wires
(4.0 cm = 0.04 m)
Substituting the values:
F/L =
(4π × 10⁻⁷ T m/A) × (8.0 A) × (5.0 A) / (2π × 0.04 m)
= 2 × 10⁻⁴ N/m
To find the force on a 10 cm (0.1 m) section of wire A:
F = (F/L) × L = (2 × 10⁻⁴ N/m) × 0.1 m = 2 × 10⁻⁵ N
Due to the parallel flow of current, the wires experience an attractive force.
Therefore, the force on the 10 cm section of wire A is 2 × 10⁻⁵ N, and it is attractive.
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 :The solenoid has the following dimensions and properties:
Length (l): 80 cm = 0.8 m
Number of turns (N): 5 layers x 400 turns/layer = 2000 turns
Diameter (D): 1.8 cm = 0.018 m
Current (I): 8 A
given,
magnetic field inside a solenoid is
B = μ₀NI / l
Where:
μ₀ denote permeability of free space (4π x 10^-7 T m/A)
N denote number of turns
I denote current
l denote length of the solenoid
Substituting the given values:
B =
(4π × 10^(-7) T m/A) × (2000 turns) × (8 A) / (0.8 m)
Calculating this gives:
B = 8π x 10^-3 T ≈ 2.512 x 10^-2 T
Therefore, the magnitude of the magnetic field inside the solenoid near its center is approximately 2.512 x 10^-2 Tesla.
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 :The torque (τ) experienced by a coil carrying current I, having N turns, and area A, placed in a uniform magnetic field B, at an angle θ to the field direction is given by:
τ = NIAB sinθ
Where:
* N = 20 turns
* I = 12 A
* A = (10 cm)² = 0.01 m²
* B = 0.80 T
* θ = 30°
Substituting the values:
τ = 20 × 12 A × 0.01 m² × 0.80 T × sin 30° = 0.96 Nm
The coil experiences a torque of 0.96 Nm
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 :a) Current Sensitivity:
Current sensitivity of a moving coil galvanometer is defined as the deflection produced per unit current.
For a galvanometer, deflection (θ) is proportional to current (I):
θ ∝ I
Or, θ = kI
where k represents the current sensitivity constant
For a galvanometer, the deflection also depends on the magnetic field (B), number of turns (N), area of the coil (A), and the spring constant (K).
θ ∝ NIBA/K
Comparing the two equations, we get:
k ∝ NIBA/K
Therefore, the ratio of current sensitivities of M2 and M1 is:
k₂/k₁ = (N₂B₂A₂)/(N₁B₁A₁)
Substituting the given values:
k₂/k₁ = (42 × 0.50 × 1.8 × 10⁻³) / (30 × 0.25 × 3.6 × 10⁻³) = 1.4
b) Voltage Sensitivity:
Voltage sensitivity is defined as the deflection produced per unit voltage.
For a galvanometer, the voltage sensitivity is given by:
θ/V = k/R
where R represents resistance of the galvanometer.
Therefore, the ratio of voltage sensitivities of M2 and M1 is:
(θ₂/V₂) / (θ₁/V₁) = (k₂/R₂) / (k₁/R₁)
Substituting the values:
(θ₂/V₂) / (θ₁/V₁) = (1.4 × 14) / (1 × 10) = 1.96
Therefore, the ratio of voltage sensitivities of M2 and M1 is 1.96.
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 = 1.5 × 10-19 C, me = 9.1×10-31 kg)
Ans :*Why the path is a circle:**
The reason a charged particle moves in a circular path when entering a magnetic field perpendicular to its velocity is due to the magnetic force acting as a centripetal force. This force constantly pulls the particle towards the center of the circular path, resulting in circular motion
**Determining the radius of the circular orbit:**
given,
magnetic force on the electron is
F = qvB
This force acts as the centripetal force, which is given by:
F = mv²/r
Equating the two forces, we get:
qvB = mv²/r
Solving for r, we get:
r = mv / (qB)
Substituting the given values:
r =
(9.1 × 10⁻³¹ kg) × (4.8 × 10⁶ m/s)
——————————
[(1.6 × 10⁻¹⁹ C) × (6.5 × 10⁻⁴ T)]
Calculating the radius:
r ≈ 4.2 × 10⁻² m
Hence, the path of the circular orbit has a radius of roughly 4.2 centimeters.
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 :Determine frequency of revolution, we must first determine the time period (T) of the circular motion.
The time period refers to the duration required to finish one full revolution or oscillation. It can be calculated as the circumference of the circle divided by the speed of the electron:
T = 2πr / v
From the previous answer, we know:
r = mv / (qB)
Substituting this into the equation for T:
T = 2πm / (qB)
(f)Frequency is the reciprocal of time period.
f =
1 qB
———- = ————–
T (2πm)
Substituting the values:
f = (1.6 × 10⁻¹⁹ C × 6.5 × 10⁻⁴ T) / (2π × 9.1 × 10⁻³¹ kg) ≈ 18.0 MHz
The frequency of the electron’s circular motion is independent of its speed. This is due to the fact that the magnetic force, which acts as the centripetal force, is directly proportional to the electron’s speed. As the speed increases, the radius of the orbit also increases, resulting in a constant frequency of revolution.
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 :Given:
(a)
Number of turns (n) = 30
Radius of the coil (r)
= 8 cm
= 0.08 m
Area of the coil (A) = πr² = 0.0201 m²
Current (I) = 6.0 A
Magnetic field strength (B) = 1 T
Angle between field lines and normal to the coil surface (θ) = 60°
Calculation:
here,
torque on a current-carrying coil in a magnetic field is given by following formula
τ = nIBA sinθ
Substituting the given values:
τ = 30 x 6.0 A x 1 T x 0.0201 m² x sin 60°
τ = 3.133 Nm
Therefore, the magnitude of the torque acting on the coil is 3.133 Nm.
(b)The equation for torque (τ = nIBA sinθ) shows that the torque on the coil depends on its area (A), not its shape. Therefore, replacing the circular coil with a planar coil of irregular shape but the same area would not change the magnitude of the torque.
