1. A metal bar spins at a constant rate in the magnetic field of the Earth as in
ID: 2107390 • Letter: 1
Question
1. A metal bar spins at a constant rate in the magnetic field of the Earth as in Figure 31.10. The
rotation occurs in a region where the component of the Earth’s magnetic field perpendicular to
the plane of rotation is 3.30 × 10^–5 T. If the bar is 1.00 m in length and its angular speed is 5.00
Ï€ rad/s, what potential difference is developed between its ends?
2. A 70-turn solenoid is 5.00 cm long and 1.00 cm in diameter and carries a 2.00-A current. A
single loop of wire, 3.00 cm in diameter, is held so that the plane of the loop is perpendicular to
the long axis of the solenoid, as illustrated in Figure P31.18 (page 1004). What is the mutual
inductance of the two if the plane of the loop passes through the solenoid 2.50 cm from one
end?
3. Two single-turn circular loops of wire have radii R and r, with R >> r. The loops lie in the
same plane and are concentric. (a) Show that the mutual inductance of the pair is M = μ0 πr 2 / 2R.
(Hint: Assume that the larger loop carries a current I and compute the resulting flux
through the smaller loop.) (b) Evaluate M for r = 2.00 cm and R = 20.0 cm
Explanation / Answer
Answering to questions 1 and 2 needs the figures.
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3. Two single-turn circular loops of wire have radii R and r, with R >> r. The loops lie in theÂ
same plane and are concentric. (a) Show that the mutual inductance of the pair is M = ÃŽ ¼0 Àr 2 / 2R.
 (Hint: Assume that the larger loop carries a current I and compute the resulting fluxÂ
through the smaller loop.) (b) Evaluate M for r = 2.00 cm and R = 20.0 cm
flux = A B = (pi r^2) * (u0 i/2R)
emf = d(flux)/dt = L (di/dt)
==> emf = (pi r^2 u0/2R) (di/dt)
==> L = pi r^2 u0/2R
b)
L = (3.1416*0.02*0.02)*(4*3.1416e-7)/2/0.20 = 3.95 * 10^-9 H
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