Problem#1
A closely wound rectangular coil of 80 turns has dimensions of 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position where it makes an angle of 37.0° with a magnetic field of 1.10 T to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf induced in the coil?
Answer:
Rotating the coil changes the anfle between it and magnetic field, which changes magnetic flux through it. This chane induces an emf in the coil.
εav = |dΦB/dt|
with ΦB = BA cos θ, (θ is angle between the normal to the loop and B), so the average emf induced in the coil is
εav = NBA[cos θf – cos θi]/Δt
εav = (80)(1.10 T)(0.250 m)(0.400 m) [|cos 00 – cos 53.00|]/0.0600 s
εav = 58.4 V
Problem#2
The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of 1.60 cm. The coil rotates in a magnetic field of 0.0750 T. What is the angular speed of the coil if the maximum emf produced is 24.0 mV?
Answer:
the maximum emf given by
εmax = NBAω
ω = εmax/NBA = (2.40 x 10-2 V)/[120 x 0.0750 T x (0.016 m)2]
ω = 10.4 rad/s
Problem#3
A flat, rectangular coil of dimensions l and w is pulled with uniform speed through a uniform magnetic field B with the plane of its area perpendicular to the field (Fig. 1). (a) Find the emf induced in this coil. (b) If the speed and magnetic field are both tripled, what is the induced emf?
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| Fig.1 |
Answer:
(a) the emf induced in this coil given by
ε = –dΦB/dt
If a flat, rectangular coil of dimensions l and w is pulled with uniform speed b through a uniform magnetic field B with the plane of its area perpendicular to the field, then the magnetic flux through the coil is does not change. Therefore, the rate of change of magnetic flux is
dΦB/dt = 0
using equation, the induced emf in the coil is
ε = 0
(b) If the speed and magnetic field are both tripled, the induced emf is zero (ε = 0)
Problem#4
A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) circumference of wire c = 2πr and area A = πr2, so A = c2/4π
ΦB = B.A = BA = (B/4π)c2
(a) the emf induced in the loop at the instant when 9.0 s have passed is
ε = –dΦB/dt = (B/2π)(cdc/dt)
at, t = 9.0 s, c = 1.650 m – (9.0 s)(0.120 m/s) = 0.570 m
so that,
ε = (0.500 T/2π)(0.570 m)(0.120 m/s) = 5.44 x 10-3 V
ε = 5.44 mV
(b) The loop and the magnetic field are as below
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| Fig.2 |
Take into the page to be thw positive direction for A.
Magnetic flux is positive. The positive flux is decreasing in magnitude dΦB/dt is negative and ε is positive.
By right hand rule, for A into the page, positive ε is clockwise.
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