Resolving vectors question. See diagram enclosed.
Answer:
the magnitude of the force that the wire will experience = 1.8 N
Explanation:
The force on a current carrying wire placed in a magnetic field is :
F = Idl × B
where:
I = current flowing through the wire
dl = length of the wire
B = magnetic field
We can equally say that :

where : sin θ is the angle at which the orientation from the magnetic field to the wire occurs = 30°
Then;

Given that:
L = 20 cm = 0.2 m
I = 6 A
B = 3 T
θ = 30°
Then:
F = 3 × 6 × 0.2 sin 30°
F = 1.8 N
Therefore, the magnitude of the force that the wire will experience = 1.8 N
Answer:
only thing I think of when I see that is 'Just Wondering'
Explanation:
Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :

where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy 
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J