Question 40: Units , Dimensions and Error Analysis

Question

Given a charge q , current I and permeability of vacuum $\mu_{\mathrm{o}^*}$. Which of the following quantity has the dimension of momentum ?

Accepted Answer

Let's look at the units (dimensions) of the given quantities:
Charge $ Q $ has units of ampere ÃÂ time $ (A \cdot T) $
Current $ I $ has units of ampere $ (A) $
Permeability of vacuum $ \mu_0 $ has units of mass ÃÂ length ÃÂ time$^{-2} $ ÃÂ ampere$^{-2} $ $ (MLT^{-2}A^{-2}) $
Momentum has units of mass ÃÂ length ÃÂ time$^{-1} $ $ (MLT^{-1}) $
We want to combine $ Q $, $ \mu_0 $, and $ I $ so that the result has the same units as momentum $ (MLT^{-1}) $.
Suppose the answer is $ Q^x\, \mu_0^y\, I^z $. Let's write out the units for this:
$ Q^x\, \mu_0^y\, I^z = (A T)^x \, (MLT^{-2}A^{-2})^y \, (A)^z $
Combine the powers of each unit:
Mass (M):      exponent = $ y $
Length (L):    exponent = $ y $
Time (T):      exponent = $ x - 2y $
Ampere (A):    exponent = $ x - 2y + z $
We want these to match the units of momentum, $ M^1L^1T^{-1} $:
So we set up:
$ y = 1 $ (so mass and length exponents match momentum)
$ x - 2y = -1 $ (time exponent)
$ x - 2y + z = 0 $ (ampere exponent cancels out)
Solve these equations:
First, $ y = 1 $.
Second, $ x - 2(1) = -1 \implies x = 1 $.
Third, $ x - 2(1) + z = 0 \implies 1 - 2 + z = 0 \implies z = 1 $.
The required combination is $ x = 1 $, $ y = 1 $, $ z = 1 $:
So, the quantity with the dimension of momentum is $ Q \mu_0 I $.

Answer

${ }$ $q I / \mu_o$

Answer

$q^2 \mu_o I$

Answer

$q \mu_0 / I$

Answer

$\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}$

Units , Dimensions and Error Analysis

Given a charge q , current I and permeability of vacuum \mu_{\mathrm{o}^*}. Which of the following quantity has the dimension of momentum ?

Choose the correct answer: