StudentAidUnit 10: Electromagnetism — Numericals
11th Class Physics · Unit 10: Electromagnetism
10.1.A positively charged particle is projected perpendicularly into a magnetic field at a speed of 1500 m s⁻¹. It experiences a force of magnitude F. At what angle θ with the field, the particle should be projected at a speed of 2000 m s⁻¹, so that it experiences the same magnitude of force?
Given
Initial velocity
v1 = 1500 m s-1
Final velocity
v2 = 2000 m s-1
Angle of projection in first case
θ1 = 90^circ
Formula
Force on charged particle
F = qvB sinθ
Force in first case
F1 = qv1 B sin θ1
Force in second case
F2 = qv2 B sin θ2
Setting F₁ = F₂
qv1 B sin θ1 = qv2 B sin θ2
Simplifying
v1 sin θ1 = v2 sin θ2
Substituting θ₁
v1 sin(90^circ) = v2 sin θ2
Working
Finding angle
sin θ2 = v1v2 = 15002000 = 0.75
Result
Required angle
θ2 = sin-1(0.75) = 48.59^circ ≈ 49^circ
10.2.Electrons are accelerated from rest through a potential difference of 15 kV in an oscilloscope. The electrons then pass through a 0.35 T magnetic field that deflects them to the desired position on the screen. Find the magnitude of the maximum force that an electron can experience.
Given
Potential difference
V = 15 kV = 15000 V
Charge
q = e = 1.6 × 10-19 C
Magnetic field
B = 0.35 T
Mass of electron
me = 9.1 × 10-31 kg
Formula
Kinetic energy from potential
12mv2 = eV
Solving for velocity squared
v2 = 2eVm
Working
Substituting values
v = sqrt{frac{2 × 1.6 × 10-19 × 15000}{9.1 × 10-31
Computing
v = 7.26 × 107 m s-1
Formula
Maximum force
F = qvB
Working
Substituting
F = 1.6 × 10-19 × 7.26 × 107 × 0.35
Result
Maximum force
F = 4.1 × 10-12 N
10.3.A square coil of side 15 cm each consists of 60 turns. Initially, it is located in a uniform magnetic field of magnitude 0.8 T such that plane of the coil is perpendicular to the field. The coil is then turned through an angle of θ = 30° in a time of 2 s. Determine the average induced emf.
10.4.A metallic rod is moving through a uniform magnetic field of 0.2 T. The emf induced across its ends is found to be 0.8 V. It is required to induce an emf of 2.4 V across its ends. How much field strength is needed for this?
Given
Initial magnetic field
B1 = 0.2 T
Initial induced emf
varepsilon1 = 0.8 V
Required induced emf
varepsilon2 = 2.4 V
Formula
Emf in first case
varepsilon1 = VB1L
Emf in second case
varepsilon2 = VB2L
Since v, L, θ remain constant
varepsilon1varepsilon2 = B1B2
Working
Solving for B₂
B2 = B1 × varepsilon2varepsilon1 = 0.2 × 2.40.8
Result
Required magnetic field
B2 = 0.6 T
10.5.A copper ring has a radius of 4.0 cm and resistance of 1.0 mΩ. A magnetic field is applied over the ring, perpendicular to its plane. If the magnetic field increases from 0.2 T to 0.4 T in a time interval of 5 × 10⁻³ s, what is the current in the ring during this interval?
Given
Radius
r = 4 cm = 0.04 m
Resistance
R = 1 mΩ = 10-3 Ω
Initial magnetic field
B1 = 0.2 T
Final magnetic field
B2 = 0.4 T
Change in field
ΔB = 0.2 T
Time interval
Δt = 5 × 10-3 s
Area of ring
A = pi r2 = 3.14 × (0.04)2 = 0.005024 m2
Formula
Induced emf
varepsilon = -N Δ PhiΔt = -ΔB pi r2Δt
Working
Substituting
varepsilon = -frac{0.2 × 3.14 × (0.04)2}{5 × 10-3
Computing
varepsilon = -0.201 V
Formula
Current
I = varepsilonR
Working
Substituting
I = frac{0.201}{10-3
Result
Induced current
I = 201 A
10.6.A coil of 10 turns and 35 cm² area is in a perpendicular magnetic field of 0.5 T. The coil is pulled out of the field in 1.0 s. Find the induced emf in the coil as it is pulled out of the field.
Given
Number of turns
N = 10
Area of cross-section
A = 35 cm2 = 35 × 10-4 m2
Magnetic field
B = 0.5 T
Time
Δt = 1.0 s
Initial magnetic flux
Phi1 = NBA = 10 × 0.5 × 35 × 10-4 = 1.75 × 10-2 Wb
Final flux (pulled out)
Phi2 = 0
Change in flux
Δ Phi = Phi2 - Phi1 = -1.75 × 10-2 Wb
Formula
Induced emf
varepsilon = -N Δ PhiΔt
Working
Substituting
varepsilon = -10 × frac{-1.75 × 10-2{1}
Result
Induced emf
varepsilon = 1.7 × 10-2 V
10.7.A proton is accelerated by a potential difference of 6 × 10⁵ volts. It then enters perpendicularly in a uniform magnetic field B = 1.0 weber m⁻². Find the radius of curvature of the path of the proton. m = 1.67 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C.
Given
Mass of proton
mp = 1.67 × 10-27 kg
Potential difference
ΔV = 6 × 105 V
Magnetic field
B = 1.0 Wb m-2
Charge
e = 1.6 × 10-19 C
Formula
Energy relation
12mv2 = eV
Solving for v²
v2 = 2eVm = frac{2 × 1.6 × 10-19 × 600000}{1.67 × 10-27
Computing
v = 1.07 × 107 m s-1
Magnetic force and radius
eVB = mv2r
Solving for r
r = mveB
Working
Substituting
r = 1.67 × 10-27 × 1.07 × 1071.6 × 10-19 × 1
Result
Radius of curvature
r = 11.2 cm
10.8.A proton enters a uniform magnetic field B = 0.300 weber m⁻² in a direction making an angle 45° with the magnetic field. What will be the radius of the circular path if the velocity of proton is 10⁴ m s⁻¹.
Given
Mass of proton
mp = 1.67 × 10-27 kg
Charge
e = 1.6 × 10-19 C
Magnetic field
B = 0.300 Wb m-2
Angle with field
θ = 45^circ
Velocity
v = 104 m s-1
Perpendicular component of velocity
v_perp = v sin θ = 104 sin 45^circ
Formula
Radius formula
r = mv_perpeB
Working
Substituting
r = 1.67 × 10-27 × 104 × sin 45^circ1.6 × 10-19 × 0.300
Result
Radius of circular path
r = 2.26 × 10-4 m
10.9.Three identical conducting rods L₁, L₂ and L₃ are moving in different planes with the same speeds v₁ = v₂ = v₃ = 2.5 m s⁻¹ as shown in the figure. The length of each rod is 60 cm. A constant magnetic field of magnitude B = 0.5 T is directed along z-axis. Find the magnitude of emf induced in each rod and indicate which end of the rod is positive. [(rod L₁) e.m.f. = 0.75 V, and end a, (rod L²) e.m.f. = 0 (Rod L₃) e.m.f. = 0].
Given
Velocity of all rods
v1 = v2 = v3 = 2.5 m s-1
Length of each rod
L1 = L2 = L3 = 60 cm = 0.6 m
Magnetic field
B = 0.5 T
Formula
Induced emf
varepsilon = vBL sin θ
For Rod 1: perpendicular motion
θ = 90^circ
Working
Rod 1 calculation
varepsilon1 = 2.5 × 0.5 × 0.6 × sin 90^circ = 0.75 V
For Rod 2: parallel motion
θ = 0^circ
Rod 2 calculation
varepsilon2 = 2.5 × 0.5 × 0.6 × sin 0^circ = 0
For Rod 3: parallel motion
θ = 0^circ
Rod 3 calculation
varepsilon3 = 2.5 × 0.5 × 0.6 × sin 0^circ = 0
varepsilon1 = 0.75 V (end a positive); varepsilon2 = 0; varepsilon3 = 0
10.10.An emf of 0.5 V is induced across the ends of a metal rod moving through a magnetic field of 0.4 T. If an emf of 1.5 V has to be induced, what field strength would be needed for that? Assume that all other factors remain the same.
Given
Initial emf
varepsilon1 = 0.5 V
Initial magnetic field
B1 = 0.4 T
Required emf
varepsilon2 = 1.5 V
Formula
Induced emf (first case)
varepsilon1 = VB1L
Induced emf (second case)
varepsilon2 = VB2L
As v, L, θ remain constant
varepsilon1varepsilon2 = B1B2
Working
Solving for B₂
B2 = B1 × varepsilon2varepsilon1 = 0.4 × 1.50.5
Result
Required field strength
B2 = 1.2 T
10.11.A charged particle moves through a velocity selector at a constant velocity of 4.9 × 10⁴ m s⁻¹ in a direction perpendicular to both E and B. If the magnetic field strength is 0.114 T, what should be the magnitude of electric field intensity so that the particle moves undeflected?
Given
Velocity
v = 4.9 × 104 m s-1
Magnetic field
B = 0.114 T
For undeflected motion, electric and magnetic forces balance
FE = FB
Formula
Electric force
FE = qE
Magnetic force
FB = qvB
Setting forces equal
qE = qvB
Working
Solving for E
E = vB = 4.9 × 104 × 0.114
Result
Electric field intensity
E = 5.65 × 103 N C-1
10.12.A current-carrying conductor PQ of length 2 m is placed perpendicularly to a magnetic field of flux density 0.5 T as shown in the figure. The resulting force on the conductor is 1 N acting into the plane of the paper. What is the magnitude and direction of the current?
Given
Length of conductor
ell = 2 m
Magnetic field
B = 0.5 T
Force on conductor
F = 1 N
Angle with field
θ = 90^circ
Formula
Force on current-carrying conductor
F = ILB sin θ
Since θ = 90°
sin 90^circ = 1
Working
Solving for current
I = FLB = 12 × 0.5
Result
Magnitude of current
I = 1 A
Direction (Fleming's left-hand rule)
From Q to P