Download Study problems – Magnetic Fields – With Solutions Not to be turned

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 × 108 𝑚
)
𝑠
⃑ = 0.5 × 10−4 𝑇 North
𝐵
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.
⃑  |𝑞||𝑣||𝐵|𝑠𝑖𝑛𝜃  𝜃 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑣 𝑎𝑛𝑑 𝐵
⃑ = 90° sin 𝜃 = 1 𝐹 = |𝑞||𝑣||𝐵|
𝐹 = 𝑞𝑣 × 𝐵
𝐹 = (1.6 × 10
𝑟=
𝑚𝑣
𝑞𝑏
−19
𝑟=
3 × 108 𝑚
𝐶)(0.12) (
) (0.5 × 10−4 𝑇) = 2.88 × 10−16 𝑁
𝑠
108 𝑚
)
𝑠
−19
−4
(1.6×10
𝐶)(0.5×10 𝑇)
(1.67×10−23 𝑘𝑔)(0.12)(3×
=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?
1
𝐾𝐸 = 2 𝑚𝑣 2 =
𝑞 2 𝐵2 𝑅2
2𝑚
𝑟=
𝐸
𝑣 = 𝐵  𝐸 = 𝑣𝐵  𝐸 = (
𝑚𝑣
𝑞𝐵
35000𝑚
𝑠
𝑟=
35000𝑚
)
𝑠
(1.67×10−27 𝑘𝑔)(
(1.6×10−19 )(0.75𝑇)
= 0.49𝑚𝑚
) (0.75𝑇) = 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)?
⃑  𝑙×𝐵
⃑ for parallel =0
𝐹 = 𝐼𝑙 × 𝐵
𝐹 = (24𝑚𝐴)(0.45𝑚)(0.5 × 10−4 𝑇) = 5.4 × 10−8 𝑁 𝑢𝑝
𝐹 = 5.4 × 10−8 𝑁 𝑢𝑝
𝐹 = 5.4 × 10−8 𝑁 𝑑𝑜𝑤𝑛
⃑ =
𝜏 = 𝐼𝐴 × 𝐵
(24𝑚𝐴)(0.045𝑚)(0.5 × 10−4 )
2.4 × 10−9 𝑁𝑚
For a circle = IAB  𝐴 = 𝜋𝑟 2  4𝑙 = 2𝜋𝑟  𝑟 =
2𝑙
𝜋
2𝑙
 𝐴 = 𝜋 ( 𝜋 )2 3 × 10−9 𝑁𝑚