\'Il T-Mobile 8:14 A quest.cns.u This print-out should have 20 questions Multipl

Question

'Il T-Mobile 8:14 A quest.cns.u This print-out should have 20 questions Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A uniform 1.1 T magnetic field points north ward the ground) with a speed of 2.7 x 10 m/s through this eld. The charge on a proton is 160x10 a) What is the magnitude of the force acting on it? Answer in units of N 002 (part 2 of 2) 10.0 points b) What is the direction of the force? . North 2. East 3. None of these 4. West 5. South 003 10.0 points An electron in a vacuum is first accelerated by a voltage of 11000 V and then enters a region in which there is a uniform magnetic fiell of 0.271 T at right angles to the direction of the electron's motion and its charge is 1.60218 x 10 electron due to the magnetic field? The mass of the electron is 9.11 x 10- kg C What is the magnitude of the force on the Answer in units of N. 004 10.0 points A proton moves perpendicularly to a mag- netic field that has a magnitude of 3.67 x 10 The charge on a proton is 1.60x 10-1 c. What is the speed of the particle if the magnitude of the magnetic force on it is 1.18x Answer in units of m/s

Explanation / Answer

1) magnetic foece f = q*(v*B)

here v = -2.7*10^7 m/s j

B = 1.1 T j

j*j = 0

so force is 0 N

2) none of these

3) 1/2mv^2 = qV

v = sqrt(2*1.60218*10^-19*11000/9.11*10^-31)

v = 6.22*10^7 m/s

f = qvB = 1.60218*10^-19*6.22*10^7*0.271 = 2.7*10^-12 N

4) v = f/qB = 1.18*10^-14/1.6*10^-19*3.67*10^-2 = 2.01*10^7 m/s

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