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Learn/ 12th Physics/ Unit 14 /Numericals

Unit 14: Electromagnetism — Numericals

12th Class Physics · Unit 14: Electromagnetism

14.1.A 20.0 cm wire carrying a current of 10.0 A is placed in a uniform magnetic field of 0.30 T. If the wire makes an angle of 40° with the direction of magnetic field, find the magnitude of the force acting on the wire.
Given
Length of wire L = 0.20 m
Current I = 10.0 A
Magnetic field B = 0.30 T
Angle θ = 40circ
Formula
Force on conductor F = IBL sin θ
Working
Substituting values F = (10.0)(0.30)(0.20) sin 40circ
= 0.39 N
F = 0.39 N
14.2.The magnetic field in a certain region is given by B = (40î - 18k̂) Wbm⁻². How much flux passes through a 5.0 cm² area loop in this region if the loop lies flat in the xy-plane?
Given
Magnetic induction vec{B} = (40hat{i} - 18hat{k}) Wbm-2
Area of loop A = 5.0 cm2 = 5.0 × 10-4 m2
Formula
Magnetic flux phiB = vec{B} · vec{A}
Working
For loop in xy-plane with normal +k phiB = (40hat{i} - 18hat{k}) · (5.0 × 10-4hat{k})
= -90 × 10-4 Wb
phiB = 90 × 10-4 Wb
14.3.A solenoid 15.0 cm long has 300 turns of wire. A current of 5.0 A flows through it. What is the magnitude of magnetic field inside the solenoid?
Given
Length of solenoid L = 15.0 cm = 0.15 m
Number of turns N = 300
Current I = 5.0 A
Permeability of free space μ0 = 4pi × 10-7 WbA-1m-1
Formula
Magnetic field in solenoid B = μ0 n I
Where n = N/L n = NL = 3000.15
Working
Substituting B = μ0 NL I = 4pi × 10-7 × 300 × 5.00.15
B = 4pi × 10-7 × 15000.15
B = 1.3 × 10-2 Wbm-2
B = 1.3 × 10-2 Wbm-2
14.4.Find the radius of an orbit of an electron moving at a rate of 2.0 × 10⁷ ms⁻¹ in a uniform magnetic field of 1.20 × 10⁻³ T.
Given
Speed of electron v = 2.0 × 107 m/s
Magnetic field strength B = 1.20 × 10-3 T
Mass of electron m = 9.11 × 10-31 kg
Charge on electron e = 1.6 × 10-19 C
Formula
Radius of circular orbit r = mveB
Working
Substituting values r = frac{9.11 × 10-31 × 2.0 × 107{1.6 × 10-19 × 1.20 × 10-3
= 9.43 × 10-2 m
r = 9.43 × 10-2 m
14.5.Alpha particles ranging in speed from 1000 ms⁻¹ to 2000 ms⁻¹ enter into a velocity selector where the electric intensity is 300 Vm⁻¹ and the magnetic induction 0.20 T. Which particle will move undeviated through the field?
Given
Electric intensity E = 300 Vm-1
Magnetic field B = 0.20 T
Formula
For undeviated motion (electric and magnetic forces balance) Ee = eVB
Simplifying V = EB
Working
Substituting V = 3000.20
= 1500 m/s
V = 1500 m/s
14.6.What shunt resistance must be connected across a galvanometer of 50.0 Ω resistance which gives full scale deflection with 2.0 mA current, so as to convert it into an ammeter of range 10.0 A?