What are the units for the right side of this equation?

  • Thread starter smithj1990
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In summary, the right side of the given equation is in units of Tesla multiplied by meters squared divided by seconds, or T*(m^2)/s. This is because the left and right sides of the equation must have matching units, and the two terms on the right side must also have matching units. The units for each of the constants in the equation must also be considered, such as the units for Tesla being 1 kg/(A*dots^2). The vector curl operation transforms the units of the B-field, and can be understood by looking at its "Usage" section on Wikipedia.
  • #1
smithj1990
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[itex]
\nabla\times\mathbf{B} = \mu_o\mathbf{J} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}
[/itex]

someone tell me what the units are for the right side of this equation?

is it T*(m^2)/s ?
 
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  • #2
smithj1990 said:
[itex]
\nabla\times\mathbf{B} = \mu_o\mathbf{J} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}
[/itex]

someone tell me what the units are for the right side of this equation?

is it T*(m^2)/s ?

The units have to be the same for the LHS and RHS of the equation, and the units have to match for the two terms on the RHS.

Can you show the units for each of the 3 terms (1 on LHS and 2 on RHS)? Start with the obvious units, and then express the composite units in terms of the fundamental mksA units...
 
  • #3
well the left hand side is T and the first part of the right hand side is F*A/m and I am really not too sure about the rest..
 
  • #4
smithj1990 said:
well the left hand side is T and the first part of the right hand side is F*A/m and I am really not too sure about the rest..

No, when you take the vector curl, what are you doing to the B field? And beyond that, what are the fundamental mksA units of a Tesla? Either your textbook should have a handy Units guide, or you can find one in the Internet. My Applied EM text by Plonus has a great Units guide inside the front cover, and a Vector Operations guide in all 3 coordinate systems inside the back cover...

And what are the units for each of the constants in the equation?
 
  • #5
well i know that 1 Tesla is 1 kg/(A[tex]\dot[/tex]s^2)

i don't really understand vector curl.. i try to go about this by the integral approach over a closed loop. can someone explain this to me.. or at least tell me if my original thought is correct?
 
  • #6
smithj1990 said:
well i know that 1 Tesla is 1 kg/(A[tex]\dot[/tex]s^2)

i don't really understand vector curl.. i try to go about this by the integral approach over a closed loop. can someone explain this to me.. or at least tell me if my original thought is correct?

Correct about the fundamental mksA units for Tesla. Where did you find that? Are the other expansions into fundamental mksA units given there for Volts, Farads, Coulombs, etc.?

For the vector curl, look at this "Useage" section of the wikipedia article for vector curl:

http://en.wikipedia.org/wiki/Curl_(mathematics )

Do you see what the operation does? How do you think the curl operator transforms the units of the B-field when you take the curl of B?
 
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What units for this equation?

The units for an equation depend on the physical quantities being measured. To determine the units, look at the variables in the equation and their corresponding units. For example, if the equation is force = mass x acceleration, the units would be Newtons (N) for force, kilograms (kg) for mass, and meters per second squared (m/s^2) for acceleration.

How do I convert units in an equation?

To convert units in an equation, use conversion factors or dimensional analysis. Conversion factors are ratios that relate different units of measurement, while dimensional analysis involves canceling out units to arrive at the desired final unit. It is important to maintain the correct units throughout the calculation to ensure accurate results.

Why are units important in scientific equations?

Units are important in scientific equations because they provide context and meaning to the numbers being calculated. They also allow for consistency and comparability in measurements and calculations. For example, knowing the units for a velocity equation (meters per second) can help us understand the speed of an object and compare it to other velocities.

What happens if I use the wrong units in an equation?

If you use the wrong units in an equation, you will get an incorrect result. This is because the units represent the physical quantities being measured, and using the wrong units will result in a different quantity being calculated. Additionally, using the wrong units can make it difficult for other scientists to understand and replicate your work.

Can different units be used in the same equation?

Yes, different units can be used in the same equation as long as they are equivalent. This means that they have the same physical meaning or can be converted into each other using a conversion factor. However, it is important to use consistent units throughout a calculation to avoid errors.

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