Question

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magnetic field is produced by horizontal coils embedded in the bottom of the channel. A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 6.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .Part A) What is the magnitude of the field at the center of the coil?Part B) What is the direction of the force on a positive ion in the water above the center of the coil?The force points west.The force points north.The force points south.The force points east.Part C) What is the magnitude of the force on an ion with a charge +e?

Answer 1

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

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

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

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) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

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

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$