Calculating the Magnitude of the Force

Themes > Science > Physics > Electromagnetism > Magnetostatics > Lorentz Force Law > Magnetohydrodynamics and the Lorentz Force Law > Calculating the Magnitude of the Force

Once you have found the direction of the force you need to calculate its MAGNITUDE. You can't directly use the standard form of the Lorentz force law since the values of Q, v and B would be very difficult to measure. So you have to convert the equation in terms of things which are easily measured:

Start with F = QvB

I = Q / t (definition of current) therefore Q = I t

v = d / t (definition of velocity)

substitute to get

F = (I t)(d / t)B simplify...

F = IdB

I is the current of charges

d is the distance traveled by the charges

B is the magnetic field (perpendicular to the direction of the current)

The boat itself needs a reasonable force to accelerate it. The following calculations are based on my construction of the boat.

Use F = I d B

I = 3 A (short circuit current between electrodes)

d = 2.7 cm (0.027 m) (distance between electrodes)

B = 0.2 T ( S.I. units: 0.2 N/(A*m) measured)

therefore force on water (force propelling the boat) is:

F = (3.0 A)(0.027 m)(0.2 N/(A*m))

F = 0.0162 Newtons

It is only natural that students in grade 12 physics don't have an intuitive idea of how much this force actually is, so to put things into perspective you ask the question:

what mass would this be if the force is due to gravity acting on that mass?

F = ma = mg therefore m = F / g

F = 0.0162 N ; g = 9.8 m/s2

m = (0.0162 kg*m/s2) / (9.8 m/s2) = 1.6 grams

is about the mass of a penny.

What acceleration does this produce on the boat?

Boat has mass 0.470 kg

F = ma therefore a = F/m = (0.0162N)/(0.47kg) = 0.04 m/ s2

What speed will the boat be going after 30 s?

a = v/t therefore v = at = (0.04m/s2)(30s) = 0.0012 m/s = 1.2 mm/s

Not fast but still noticeable!

Information provided by: http://www.physics.ubc.ca