Problem 30.19 Enhanced - with Feedback Part G The magnetic field in the figure i

Question

Problem 30.19 Enhanced - with Feedback Part G The magnetic field in the figure is decreasing at the rate 0.4 T/s. (Figure 1) What is the magnitude of the acceleration of a proton at rest at point d? Express your answer to two significant figures and include the appropriate units. You may want to review ( Dages 852-855) Submit My Answers Give Up Incorrect, Try Again, 5 attempts remaining Part H What is the direction of the acceleration of a proton at rest at point d? Figure 1 |ofl downward horizontally to the left horizontally to the right upward 2.0 cm Submit My Answers Give Up 1 cm 1 cm 1 cnm Correct Here we learn how to determine the direction of the acceleration of a proton at rest in a solenoid with a changing magnetic field

Explanation / Answer

30.19

given rate of change of magnetic filed

dB/dt = -0.4

radius of loop a= 0.02 m

E. for a proton at rest at point c

emf, V = -pi*0.01^2*d(B)/dt = pi*0.01^2*0.04*(-1) = 0.000004 *piV

Electric field E = V/2*pi*0.01 = 0.0002 V/m

hence a = qE/m = 1.6*10^-19*E/1.6*10^-27 = 20000 m/s/s

F. as magnetic field into the paper is decreasing, the proton will move in the direction of current that increases magnetic filed into the page

hence clockwise

hence force is downwards

G. acceleration at point d = a

V = -pi*a^2(dB/dt)

EMF = V/2*pi*a = -a(dB/dt)/2

a = qE/m = -qa(dB/dt)/2m = 400000 m/s/s

H. from part F, direction of accelration = downwards

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