Question

The dimensions of $\frac{mB}{kT}$ where $m$ is magnetic moment, $B$ is magnetic flux density, $k$ is Boltzmann constant and $T$ is temperature are ________.

• A. $ML^{-1}T^{-1}$
• B. $ML^{2}T^{-1}$
• C. $MLT^{-1}$
• D. $ML^{-2}T$
• E. $M^{0}L^{0}T^{0}$

Hint:

Ratios of similar quantities are dimensionless.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
We need to find the dimensions of the given physical quantity. This involves finding the dimensions of each component (m, B, k, T) and then combining them according to the formula. A dimensionless quantity has dimensions \(M^0L^0T^0\).
Step 2: Key Formula or Approach
We need the dimensional formulas for each term:
1. Magnetic moment (m): The potential energy U of a magnetic dipole in a magnetic field B is \(U = -mB\). Energy has dimensions of \(ML^2T^{-2}\). The unit of B is Tesla. Alternatively, we can use the definition \(m = IA\) (current \(\times\) area). The dimension of current (I) is A, and area is \(L^2\). So, \([m] = AL^2\). 2. Magnetic flux density (B): From the Lorentz force formula \(F = qvB\), we have \(B = F/(qv)\). \([F] = MLT^{-2}\), \([q] = AT\) (charge = current \(\times\) time), \([v] = LT^{-1}\). \([B] = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{ALT} = MT^{-2}A^{-1}\). 3. Boltzmann constant (k): From the equipartition theorem, the average kinetic energy of a molecule is proportional to temperature, e.g., \(E = \frac{3}{2}kT\). So, \(k = E/T\). \([E]\) (Energy) = \(ML^2T^{-2}\), \([T]\) (Temperature) = K. \([k] = \frac{ML^2T^{-2}}{K} = ML^2T^{-2}K^{-1}\). 4. Absolute temperature (T): The dimension is K.
Step 3: Detailed Explanation
1. Find the dimensions of the numerator (mB).
The product mB has the dimensions of energy (U = mB).
So, \([mB] = [Energy] = ML^2T^{-2}\).
(Let's verify this using the individual dimensions: \([m][B] = (AL^2)(MT^{-2}A^{-1}) = ML^2T^{-2}\). This confirms our approach.)
2. Find the dimensions of the denominator (kT).
The product kT also has the dimensions of energy (\(E = kT\)).
So, \([kT] = [Energy] = ML^2T^{-2}\).
(Let's verify this using the individual dimensions: \([k][T] = (ML^2T^{-2}K^{-1})(K) = ML^2T^{-2}\). This is also correct.)
3. Combine the dimensions.
Now we find the dimensions of the entire expression \(\frac{mB}{kT}\).
\[ \left[\frac{mB}{kT}\right] = \frac{[mB]}{[kT]} = \frac{[Energy]}{[Energy]} = \frac{ML^2T^{-2}}{ML^2T^{-2}} \] \[ = M^{1-1}L^{2-2}T^{-2-(-2)} = M^0L^0T^0 \] Step 4: Final Answer
The expression \(\frac{mB}{kT}\) is a ratio of two energies, making it a dimensionless quantity. Its dimensions are \(M^0L^0T^0\).