A proton of mass 1.67x10-27 kg and charge 1.6x10-19 C is moving in the x-directi

Question

A proton of mass 1.67x10-27 kg and charge 1.6x10-19 C is moving in the x-direction with a constant speed of 104 m/s in a region of space containing a constant electric field and a constant magnetic fic the magnetic field is 10 T in the y-direction, (a) what is the magnitude and direction of the electric field that will keep the particle moving in a straight line at constant speed? (b) Describe the subsequernt motion of the particle if the electric field is suddenly turned off. (c) What is the maximum distance th pr field. If oton would deviate from its initial position when the electric field is turned off?

Explanation / Answer

a) Force due to electric field has to be equal and opposite to magnetic force

FB - magnetic force

FB= q(v X B)= q x (104 i x 10 j) = q105 k N (along +z axis)

Thus, electric force has to act along '-z' axis and be equal to q105N

Thus, the electric field is 105 N/C

b) if the electric field is switched off the particle will follow a curvilinear motion in the x-z plane and bend towards positive z-axis

c)

the proton undergoes a circular motion

thus, m*v2/r = q*v*B

therefore, r= m*v/q*B

= 10-5 m

question 2

a) initial rate of increase of current in a RL circuit

as, V(t) = VR + VL = I x R + L di/dt

as initially I = 0,

therefore, L di/dt = V(t)

di/dt = V(t)/L = 12/3 = 4 A/s

b) when I = 1A

12 = 1 x 7 + L x di/dt

5/L = di/dt

therefore, di/dt= 5/3 = 1.67 A s

c) Instantaneous current can be easily shown to be equal to

I(t) = (V/R) x (1-e-Rt/L)

= (12/7) x( 1-e-7*0.2/3)

= 1.714 x (1-0.626)=0.641 A

d) steady state current = V/R = 12/7 = 1.714 A

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