Answer
a. Magnetic south pole
b. 3.86 A
c. $8.50\times10^{-2}N$
Work Step by Step
a. Use the right-hand rule to show that the magnetic field points up. The top pole face is a magnetic south pole.
b. Use equation 20–2. The length of the wire in the field is the diameter of the pole face (i.e., the approximate length where the field envelops the wire).
$$F_{max}=I\mathcal{l}B $$
$$I=\frac{F_{max}}{\mathcal{l}B}=\frac{8.50\times10^{-2}N}{(0.100m)(0.220T)}=3.86A $$
c. Now the wire is tipped – assume it is tipped above the horizontal (the answer is the same if we make the assumption that it is tipped below). The angle between the wire and the magnetic field changes to 80 degrees. But there is now more wire inside the field: the length changes from $\mathcal{l}$ to $\frac{\mathcal{l}}{cos10^{\circ}}$.
Use equation 20–1 to calculate the force.
$$F_{old}=I\mathcal{l}Bsin90^{\circ}$$
$$F_{new}=I\frac{\mathcal{l}}{cos10^{\circ}}Bsin80^{\circ}$$
The force does not change; it is still $8.50\times10^{-2}N$.
