Question

Problem 20.58 (Multistep) In the figure below, the center of a large bar magnet is z = 22 cm from a thin plate of high-resistance material 12 cm long, 2.5 cm high, and 0.1 cm thick that is connected to a 240 V power supply whose internal resistance is negligible. The bar magnet is perpendicular to the plate. Voltmeter AV = +1.2E-5 V d=0.1 cm L = 12 cm h=2.5 cm Ammeter Power supply I=?A K = 240 V The mobility of charge carriers in the thin plate is 3.1 x 10-4 (m/s)/(V/m), and the number density of the singly charged charge carriers is 3.6 x 1023 m-3. A voltmeter is connected precisely vertically across the thin plate and reads +1.2 x 10-5 V. A low-resistance ammeter is in series with the rest of the circuit.

What is the magnetic dipole moment of the bar magnet?

Answer 1

Answer:

Girnen NH = 1.2 x 105W h = 306 x 1023 m3. l = 2.5cm = 0.025m -19 'electione (-6x10 'I resing hall voltage equation, VH = I. B ( nolie 102x105 - I XB (3.6x1023) (0.025) (1.6x1019) I-XB = 0.173 *10' I XB = 0.0173 using v Exb equation, 240V - Ex 0.12 E = 240 - 2000 W/m 0-12 As we know - mobility Me = Vd where vd drift velocity 301x104 = wd 2000 ² Vd = 0.62 m/s using equation for current I= n. e. And - 3.6 X 2023) (1-6x1019)(0.142.5)(0.62) I I. 0.893 - resing I XB=0.0173 B = 0.0173 =o.0194 Tesla. 0.893 - Magnetic dipole moment Ballo (2 m = B x 4 TX 83 = 0·0194= 10 x 2m > m= 0.01944 (0.22 = 1032.856 ~ 1033 AM 2x Ho. (0.223 2x107 (Answer)

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