Answer :
A) [tex]1.36\cdot 10^{-4}T[/tex]
The magnetic field at the center of a coil of N turns is given by
[tex]B=\frac{\mu_0 N I}{2R}[/tex]
where
I is the current in the coil
N is the number of turns
R is the radius of the coil
Here we have
I = 6.5 A is the current in the coil
N = 100 is the number of turns
[tex]R=\frac{6.0 m}{2}=3.0 m[/tex] is the radius of the coil
Substituting,
[tex]B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T[/tex]
B) The force points north
The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:
- Index finger: direction of motion of the ion --> towards east
- Middle finger: direction of magnetic field --> downward
- Thumb: direction of the force --> towards north
So, the force points north.
C) [tex]3.26\cdot 10^{-23}N[/tex]
The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is
[tex]F=qvB[/tex]
where
q is the charge of the particle
v is the velocity
B is the magnitude of the magnetic field
In this case, we have
[tex]q=+e=1.6\cdot 10^{-19} C[/tex] is the charge
[tex]v=1.5 m/s[/tex] is the velocity
[tex]B=1.36\cdot 10^{-4}T[/tex] is the magnetic field strength
Substituting,
[tex]F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N[/tex]