1. A proton in a particle accelerator has a speed of 5.0x10^6 m/s in the positiv
ID: 1433840 • Letter: 1
Question
1. A proton in a particle accelerator has a speed of 5.0x10^6 m/s in the positive x-direction. It passes through a 0.4- T magnetic field which has an angle of 30 degrees towards the positive y-direction from the protons direction of motion. What is the magnitude of the force on the proton?
A. 2.0x10^6 N B. 1.6x10^-13 N C. 3.2x10^-13 N D. 0.8x10^-13 N E. 2.8x10^-13 N
2. In what direction is that force?
A. 30 degrees in positive y-direction from the positive x-axis
B. 60 degrees in the negative y-direction from the positive x-axis
C. 60 degrees in the positive y-direction from the negative x-axis
D. The negative z-direction
E. The positive z-direction
3. Three equally-spaced, parrallel wires carry identical currents as shown below...
X Y Z
What is the direction of the force on wire Z?
A. Up B. Down C. Left D. Right E. The force is zero
4. What must be the radius of a cirrcular loop of wire so that the magnetic field at its center is 1.8x10^-4 T when the loop carries a current a 12A.
A. 15um B. 2.16m C. 1.33cm D. 1.33um E. 2.16mm
5. The magnetic field produced by the solenoid in an MRI machine designed for measurements on human bodies has a field strength of 7.0 T and the current in the solenoid is a very large 200 A! What is the number of turns per meter length of this solenoid if it is 1.5m in diameter and 3.0m in length?
A. 84000 B. 2800 C. 14000 D. 1400 E. 28000
Explanations Please!!!
Explanation / Answer
1.
magnetic force F = qvB sin theta
F = 1.6 e -19 * 5e 6 * 0.4 * sin 30
F = 1.6*10^-13 N , option B
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2. V is along x axis, B is at 30 deg from y direction
so F will be along
E. The positive z-direction
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3. to the left or toWards Y
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4. magnetic field B = uo i/2r
radius r = uoi/2B
r = (4*3.14 e-7* 12)/(2 * 1.8 e -4)
r = 0.0416 m ( no corect option )
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5. magnetic field due to solonoid is B = uo n i
n is no. of turns per unit length
so
n = (B/uo i)
n = (7/(200 * 4*3.14e -7)
n = 28000 turns option E it is
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