HW 08 Chapter 27 1. An alpha particle (2 protons attached to 2 neutrons) has vel
ID: 2037326 • Letter: H
Question
HW 08 Chapter 27 1. An alpha particle (2 protons attached to 2 neutrons) has velocity -(9i-12x 104 m and is at a location where the magnetic field is B-(-1.8 i + 1.2 j) T. Calculate the magnitude and direction of the force on the alpha particle. 2. The force on a wire is a maximum of 7.5x10 N when placed between the pole faces of a magnet. The current through the wire flows horizontally to the right and the magnetic field is vertical. The wire is observed to jump out of the page due to the magnetic force. a) Sketch the wire and the magnet (both south and north pole). Mark the direction of the current, magnetic field and the force on your diagram. What type of the magnetic pole is the top pole face (N or S)? b) If the pole faces have a diameter of 10 cm, estimate the current in the wire if the field is 0.22 T Hint: How is the diameter of pole faces related to the length of the wire. c) If the wire is tipped so it makes an angle of 10° with the horizontal, what force will it now feel? Give magnitude and direction. 3. A beam of electrons (each electron having kinetic energy K) exits a thin "window of foil at the end of a particle accelerator's vacuum tube. The beam is aimed at a wall that is a distance d from where the particles exit the window (see the figure). a) What is the minimum magnetic field that needs to be applied to bend the electrons so they never hit the wall? Express your answer in terms of me, e, K, and d. b) What is the direction of the applied magnetic field? c) Under the given magnetic field, the electrons would keep rotating. What is the frequency of oscillations?Explanation / Answer
1.
Magnetic force is given by:
F = q*VxB
q = +2e
V = (9 i - 12 j)*10^4 m/sec
B = (-1.8 i + 1.2 j) T
VxB = (9 i - 12 j)X(-1.8 i + 1.2 j)
VxB = 9*1.2*(ixj) + 12*1.8*(jxi)
(Since ixi = 0 = jxj & ixj = k & jxi = -k)
VxB = 10.8 k - 21.6 k
VxB = -10.8 k
Now
F = 2*e*(-10.8*10^4) k
F = -2*1.6*10^-19*10.8*10^4 k
F = -(3.456*10^-14 N) k
|F| = 3.456*10^-14 N
Direction = -k = into the page
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