Question

If \( B \) is magnetic field and \( \mu_0 \) is permeability of free space, then the dimensions of \( \frac{B}{\mu_0} \) is:

• A. \( \text{MT}^{-2} A^{-1} \)
• B. \( \text{L}^{-1} A \)
• C. \( \text{LT}^{-2} A^{-1} \)
• D. \( \text{ML}^2 T^{-2} A^{-1} \)

Hint:

The dimension of \( B/\mu_0 \) can be derived from the relationship of magnetic field with current and charge density.

Answer 1

The Correct Option is B

To find the dimensions of the expression \( \frac{B}{\mu_0} \), where \( B \) is the magnetic field and \( \mu_0 \) is the permeability of free space, we must first consider their fundamental dimensional formulas:

The magnetic field \( B \) has the dimensions given by: \([B] = \text{MT}^{-2}A^{-1}\)

The permeability of free space \( \mu_0 \) has the dimensions: \([\mu_0] = \text{MLT}^{-2}A^{-2}\)

Now, calculate \(\frac{B}{\mu_0}\):

\(\frac{B}{\mu_0} = \frac{\text{MT}^{-2}A^{-1}}{\text{MLT}^{-2}A^{-2}}\)

This simplifies to:

\(\frac{B}{\mu_0} = \frac{\text{M}^{1-1}\text{T}^{-2+2}\text{A}^{-1+2}}{\text{L}^{1}}\)

Resulting dimensions are: \(\text{L}^{-1}\text{A}^{1}\)

Therefore, the dimensions of \(\frac{B}{\mu_0}\) are \(\text{L}^{-1}A\).

Answer 2

Magnetic field (B): The dimensional formula of \( B \) is \[ [B] = [M^1 L^0 T^{-2} A^{-1}]. \]

Permeability of free space (μ₀): From the relation \[ B = \mu_0 H, \] where \( H \) (magnetic field intensity) has the dimension \[ [H] = [A L^{-1}], \] we get \[ [\mu_0] = \frac{[B]}{[H]} = \frac{[M T^{-2} A^{-1}]}{[A L^{-1}]} = [M L T^{-2} A^{-2}]. \]

Now, find dimensions of \( \dfrac{B}{\mu_0} \): \[ \frac{[B]}{[\mu_0]} = \frac{[M T^{-2} A^{-1}]}{[M L T^{-2} A^{-2}]} = [L^{-1} A^1]. \]

Answer

✅ Option 2: \( L^{-1} A \)