An electron (charge = -1.6 times 10^-19 C) is moving at 3.0 times 10^5 m/s in th

Question

An electron (charge = -1.6 times 10^-19 C) is moving at 3.0 times 10^5 m/s in the positive x direction. A magnetic field of 0.80 T is in the positive z direction. The magnetic force on the electron is: A) 0 N B) 4.5 times 10^-14 N in the positive z direction C) 4.5 times 10^-14 N in the negative z direction D) 4.5 times 10^-14 N in the positive y direction E) 4.5 times 10^-14 N in the negative y direction A charged capacitor is being discharged through a resistor. At the end of one time constant the charge has been reduced by (1 - 1/e) = 63% of its initial value. At the end of two time constants the charge has been reduced by what percent of its initial value? A) 82% B) 86% C) 100% D) between 90% and 100% E) need to know more data to answer the question

Explanation / Answer

According to the given problem,

4.) Magnetic Force on the particle
=q(V cross B)
=-1.6*10-19 (3*105*0.8) (i cross k)
=3.84 *10-14 j
j gives +ve y direction.
so, D is the answer. [Options are given worng all of them should be 3.84*10-14]

5.) As given that (1-1/e) = 63%
charge on capacitor=Q0 * (e)^(-t/to)
for t=2to
charge=Qo/(e^2)
so charge reduced =1-1/(e^2)=86% Answer: B

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