number 2 Arthur Jeffries finds that a proton moving with a velocity of 1.20 time
ID: 1450640 • Letter: N
Question
number 2
Arthur Jeffries finds that a proton moving with a velocity of 1.20 times 10^6 m/s in a particular uniform magnetic field experiences a force of 4.61 times 10^-17 N. If the direction of the velocity vector is at a right angle to the direction of the magnetic field, calculate the magnitude of the magnetic field and express your answer in Gauss (1 T = 10^4 G). You are working on a routine to stake a claim as World's Greatest Magician. Magicians love levitation, and when you can levitate something and show there are "no strings attached" you've really hit magical pay dirt. You are going to cheat a little and levitate a doll instead of an actual person, and have that doll resting on a thick wire through which you will pass a current. With current in the wire, levitation! Your doll plus wire have a mass per unit length of 0.45 kg/m and you use some serious magnets to produce a magnetic field of 1.25 T^2 (see figure). What current must you have in the wire to balance the magnetic force with the gravitational force acting on the wire, and which direction should the current go (left or right in the figure)?Explanation / Answer
let,
mass per unit length, m/l=0.45 kg/m
magnetic field, B=1.25 T
use,
F=i*l*B
m*g=i*l*B
===>
i=m*g/(l*B)
i=(m/l)*g/B
i=(0.45)*9.8/1.25
i=3.53 A
the direction of current is left side
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