Study problems – Magnetic Fields – With Solutions
Not to be turned in for grades!
Question 1 (1 point)
Draw the electric field lines for a magnetic dipole. How does it compare with the electric dipole?
Solution: The magnetic field lines of a magnetic dipole have a similar shape to the electric field lines of an electric dipole. They are different in that:
1) The source of an electric field is electric charge. The source of magnetic field is current loops (or moving electric charge).
2) Electric field lines start and stop at charges. Magnetic field lines are continuous (and pass through the middle of the magnet in the case of the bar magnet).
Question 2 (3 points)
A proton is moving at 12% of the speed of light in the direction which is 20 degrees up from west. It passes through the earth’s magnetic field which points due north with a strength of 0.5 x 10-4 T. What is the resultant force on the proton? What will the radius of curvature of its path be?
v=(0.12×3×108ms)
B=0.5×10-4T North
First, determine the direction of the force. Use the right hand rule. Place your thumb in the direction of travel of the proton (20 degrees up from west). Put your fingers in the direction of the Magnetic field (North). Your palm will then face in the direction of the Force. Force is 20 degrees West of Down.
F=qv×B è qvBsinθ è θ between v and B=90° èsinθ=1 èF=qvB
F=1.6×10-19C0.123×108ms0.5×10-4T=2.88×10-16N
r=mvqb è r=(1.67×10-23kg)(0.12)(3×108ms)(1.6×10-19C)(0.5×10-4T)=7500m
Question 3 (3 points)
A cyclotron is used to accelerate protons to a velocity of 35,000 m/s. If the magnetic field for the cyclotron is 0.75 Tesla, how large does the cyclotron have to be? If the protons are directed from the cyclotron to a velocity selector with the same magnetic field, what electric field is needed for the protons to pass through the selector?
KE=12 mv2=q2B2R22m è r=mvqB è r=(1.67×10-27kg)(35000ms)(1.6×10-19)(0.75T)=0.49mm
v=EB è E=vB è E=35000ms0.75T=26000V/m
Question 4 (3 points)
A wire loop is bent into the shape of a square with each side of length 4.5 cm. The loop is placed horizontally on a tabletop with two of the sides oriented north/south and two of the sides oriented east/west. A battery is connected so that a current of 24 mA is produced around the loop; the current flows in the clockwise direction looking from the top. What is the force produced by the earth’s magnetic field on each section of current-carrying wire? What is the overall torque on the loop? What would the torque be if the same length of wire were bent into a circle instead of a square (assuming the same current)?
F=Il×B è l×B for parallel =0
F=24mA0.45m0.5×10-4T=5.4×10-8N up
F=5.4×10-8N up
F=5.4×10-8N down
τ=IA×B=(24mA)(0.045m)(0.5×10-4)2.4×10-9Nm
For a circle = IAB è A=πr2 è 4l=2πr è r=2lπ è A=π (2lπ)2 è3×10-9Nm