Varying magnetic field
1. State the basic characteristics of a non-stationary magnetic field.
Solution:
A non-stationary magnetic field is a time-varying magnetic field. The source of this field is a moving permanent magnet or electromagnet, a stationary conductor with a time-varying current, or a moving current-carrying conductor.
2.Calculate the magnetic flux through a rectangular loop with dimensions a = 4cm, b = 5cm in a magnetic field with induction B = 1.1T, if the plane of the loop forms an angle β = 300 with the direction of induction.
Solution:
Analysis:
S = a·b = 4cm·5cm = 20cm2 = 20·10–4m2, B = 1.1T, α = 900 - β = 600
The magnetic flux through the rectangular loop is φ = 1.1·10–3Wb.
3.What is the magnetic induction B if a circular area with radius r = 5cm is penetrated by a magnetic flux of 4·10–2Wb? The area is perpendicular to the field lines. (β = 900)
Solution:
Analysis:
The magnetic induction is B = 5.1T.
4.What is the radius of a circular coil whose axis forms an angle α = 300 with B = 5.89T, if a magnetic flux of 4·10–2Wb passes through the coil?
Solution:
Analysis:
The radius of the circular coil is r = 5cm.
5.In which case will the induced voltage in a conducting loop be greater? If the magnetic flux decreases from 1Wb to zero in 0.5s, or if it increases from zero to 1Wb in 0.1s? What will be the polarity of the induced voltage?
Solution:
In case b.) Ue will be 5 times greater and will have the opposite polarity.
6.A horizontal conductor 2m long was released at time 0s and fell freely in a plane perpendicular to the north-south direction (B = 5·10–5T, β = 900). Determine the induced voltage at the ends of the conductor at time 5 seconds.
Solution:
Analysis:
The induced voltage at the ends of the conductor is Ue = –5mV.
7.Determine the self-inductance of a loop if a change in current of ΔI = 0.1A produces a magnetic flux Φ = 4·10-5Wb through the loop’s cross-section.
Solution:
Analysis:
The self-inductance of the loop is L = 4·10–4H.
8.How many turns does a coil of length 30cm have if a current I = 0.5A produces a magnetic field of intensity 833A·m-1 inside the coil?
Solution:
Analysis:
The coil has 500 turns.
9.How many turns must a cylindrical coil have in order to induce an average voltage of 10V, if the magnetic flux in its core changes from 0.024Wb to 0.056Wb in 0.32 seconds?
Solution:
Analysis:
The cylindrical coil must have N = 100 turns.
10.Calculate the self-inductance of an air-core coil with 1000 turns, length 20cm, and diameter 2.0cm. What is the energy of the magnetic field if the coil carries a current of 2.5A?
Solution:
Analysis:
The self-inductance of the coil is L = 2mH, and the energy of the magnetic field is Em = 6.25mJ.