S pole i above dipole magnet v n pole i below if this

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S pole I above dipole magnet v N pole I below If this loop is below the bar magnet, the magnetic flux through it points downward and increases in magnitude as the leading north pole of the bar magnet heads to the loop and therefore the current induced in the wall at this loop must produce upward mag- netic fields to counteract this change of flux. Such magnetic fields require currents that flow counterclockwise when viewing the falling bar magnet from above as shown by the bottom loop in the figure. For the loop above the bar magnet the mag- netic field is from the falling south pole which then points downward and decreases in mag- nitude during the bar magnet’s fall. There- fore the current induced in the wall at this loop must produce downward magnetic fields to counteract the decreasing magnetic flux through this loop. Such an induced magnetic field requires a current that flows clockwise when viewing the bar magnet from above as shown in the top curve in the figure. An alternative explanation is that the loop below the bar magnet must produce a mag- netic moment to repel the falling bar magnet to try to prevent the bar magnet from in- creasing the magnetic flux through the loop. The required direction of the magnetic mo- ment must then be upward (which repels a falling north pole), as produced by a counter- clockwise current viewed from above. Alter- natively, when the falling bar magnet is below the loop, the induced magnetic moment of the loop must attract the south pole of the falling bar magnet (which requires the direction of the magnetic moment to be downward) as produced by a clockwise current when viewed from above. 018 (part 1 of 3) 10.0 points A metal strip 2 . 8 cm wide and 0 . 06 cm thick carries a current of 22 A in a uniform magnetic field of 1 . 6 T. The Hall voltage is measured to be 4 . 29 μ V. The charge on an electron is 1 . 6 × 10 - 19 C. vector B 2 . 8 cm 0 . 06 cm I a b Calculate the drift velocity of the electrons in the strip. Correct answer: 0 . 0957589 mm / s. Explanation: Let : V H = 4 . 29 μ V = 4 . 29 × 10 - 6 V , B = 1 . 6 T , and
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karna (pk4534) – HW 08 – li – (59050) 11 w = 2 . 8 cm = 0 . 028 m . The Hall voltage as a function of the drift velocity of the electrons in the strip is V H = v d B w , so v d = V H B w = 4 . 29 × 10 - 6 V (1 . 6 T) (0 . 028 m) · 10 3 mm 1 m = 0 . 0957589 mm / s . 019 (part 2 of 3) 10.0 points The charge on the electron is 1 . 6 × 10 - 19 C. Find the number density of the charge car- riers in the strip. Correct answer: 8 . 54701 × 10 28 m - 3 . Explanation: Let : I = 22 A , t = 0 . 06 cm = 0 . 0006 m , and q = 1 . 6 × 10 - 19 C . The current is I = n A q v d n = I A q v d = I w t q v d = 22 A (0 . 028 m) (0 . 0006 m) × 1 (1 . 6 × 10 - 19 C) (9 . 57589 × 10 - 5 m / s) = 8 . 54701 × 10 28 m - 3 . 020 (part 3 of 3) 10.0 points Is point a or point b at the higher potential? 1. V a < V b 2. V a = V b 3. V a > V b correct Explanation: Apply a right-hand rule to I vector and vector B to conclude that positive charge will accumulate at a and negative charge at b , so V a > V b . 021 (part 1 of 2) 10.0 points When the coil of a motor is rotating at max- imum speed, the current in the windings is 4 A. When the motor is first turned on, the current in the winding is 7 A and the voltage is 113 V.
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