Hi there!
a)
We can use the equation for the magnetic dipole moment to solve for the number of turns:
[tex]\mu_m = NIA\vec{n}[/tex]
[tex]\mu_m[/tex] = Magnetic dipole moment (0.194 Am²)
N = Number of loops (?)
A = Area of loop (4.0 cm²)
[tex]\vec{n}[/tex] denotes the area vector, or the normal line perpendicular to the area.
We first need to convert cm² to m² using dimensional analysis.
[tex]4.0 cm^2 * \frac{0.01m}{1 cm} * \frac{0.01 m}{1cm} = 0.0004 m^2[/tex]
Rearranging the equation to solve for 'N':
[tex]N = \frac{\mu_m}{IA}\\\\N = \frac{0.194}{(8.8)(0.0004)} = \boxed{55.11 \text{ turns}}[/tex]
**Since we cannot have part of a turn, the coil has about 55 turns.
b)
For this, we can use the Right-Hand-Rule for current. Looking at the coil from the left with your curled fingers going around the coil with the fingertips pointing through and to the left in the direction of the magnetic moment, your thumb points in the COUNTERCLOCKWISE direction.
c)
Now, let's use the equation for the torque produced by a magnetic field:
[tex]\tau = \mu_m \times B[/tex]
This is a cross-product, but since our magnetic field is perpendicular to the magnetic moment, we can disregard it.
Plugging in the values for the magnetic moment and the magnetic field:
[tex]\tau = 0.194 * 45 = \boxed{8.73 Nm}[/tex]
d)
Using the other RHR (current, field, force), the coil will spin about its vertical axis in the field. In more detail, if you look at the coil from the left-hand side with its opening towards you, from this perspective, the left of the coil will come towards you, and the right side of the coil will move away.
