In this circuit, two batteries, one with a potential difference of 10 V and the

Question

In this circuit, two batteries, one with a potential difference of 10 V and the other with a potential difference of 20 V. are connected in series across a resistance of 90 omega. The power delivered by the 10-V battery is approximately A) 5.4 W B) 15 W C) 10W D) 3.3 W E) 6.7 W An alpha particle with a charge of 2e (e = 1.6 times 10-^19 C) and a mass of 4(1.66 times 10^-27 kg) is moving at right angles to a uniform magnetic field of 1.20 T. The radius of curvature of the track of the panicle is 0.20 m. What is the momentum of the alpha particle? The question cannot be answered because the speed of the particle is not given. 7.7 times 10^-20kg m/s 3.1 times 10^-19kgm/s 0.77 kg m/s 4.6 times 10^-21 kg m/s A wire of radius 0.6 cm carries a current of 10 A that is uniformly distributed over its cross section. Calculate the magnetic field strength at r = 0.2 cm divided by that at r = 4 cm. A) 1.0 B) 0.75 C) 0.50 D) 2.0 E) 3.0

Explanation / Answer

1)Applying KVL to the loop,

20 +10 -90I = 0   => I = 0.11A

W=VI = 10*0.33 = 3.3W

2) In the magnetic field,

Fc = Fe

mv^2/r = qvB

mv = qBr = (2*1.6*10^-19)*(1.20)*(4*1.66*10^-27) = 7.7*10^-20 kg*m/s

3) Use equation,

B= [(uoI)/(2 r)]*(r^2/R^2)

Where R be the distance from the center of the wire to the surface of the wire and r to be the distance between the point and the center of the wire.

B0.002= [(4*10^-7*10)/(2*0.002)]*(0.002^2/0.006^2) = 0.00011

B0.004= [(4*10^-7*10)/(2*0.004)]*(0.004^2/0.006^2) = 0.00022

B0.002/ B0.004 = 0.00011/0.00022 = 1/2

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