If B is magnetic field and 0 is permeability of free space, then the dimensions of B 0 is:

Given: B is the magnetic field and 0 is the permeability of free space. Find: The dimensions of B 0. The dimensional formula of magnetic field is [B] = [M T^-2 A^-1] The dimensional formula of permeability of free space is [ 0] = [M L T^-2 A^-2] Now, [B][ 0] = [M T^-2 A^-1][M L T^-2 A^-2] So, [B][ 0] = [L^-1 A] Therefore, the dimensions of B 0 are L^-1 A. The correct option is B. Given: B is the magnetic field and 0 is the permeability of free space. Find: The dimensions of B 0. Use the relation B = 0 H where magnetic field intensity H has dimensions [H] = [A L^-1] Hence, [ 0] = [B][H] = [M T^-2 A^-1][A L^-1] = [M L T^-2 A^-2] Now, [B][ 0] = [M T^-2 A^-1][M L T^-2 A^-2] Cancelling common powers of M and T, [B][ 0] = [L^-1 A] Therefore, the dimensions of B 0 are L^-1 A, so the correct option is B.

If B is magnetic field and 0 is permeability of free space, then… — Solution | JEE Physics

MCQEasyJEE 2025Dimensions & Dimensional Analysis

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

A
$MT^{-2} A^{-1}$
B
$L^{-1} A$
C
$LT^{-2} A^{-1}$
D
$ML^2 T^{-2} A^{-1}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $B$ is the magnetic field and $\mu_0$ is the permeability of free space.

Find: The dimensions of $\frac{B}{\mu_0}$ .

The dimensional formula of magnetic field is

[B] = [M T^{-2} A^{-1}]

The dimensional formula of permeability of free space is

[\mu_0] = [M L T^{-2} A^{-2}]

Now,

\frac{[B]}{[\mu_0]} = \frac{[M T^{-2} A^{-1}]}{[M L T^{-2} A^{-2}]}

So,

\frac{[B]}{[\mu_0]} = [L^{-1} A]

Therefore, the dimensions of $\frac{B}{\mu_0}$ are $L^{-1} A$ . The correct option is B.

Using $$B = \mu_0 H$$

Given: $B$ is the magnetic field and $\mu_0$ is the permeability of free space.

Find: The dimensions of $\frac{B}{\mu_0}$ .

Use the relation

B = \mu_0 H

where magnetic field intensity $H$ has dimensions

[H] = [A L^{-1}]

Hence,

[\mu_0] = \frac{[B]}{[H]} = \frac{[M T^{-2} A^{-1}]}{[A L^{-1}]} = [M L T^{-2} A^{-2}]

Now,

\frac{[B]}{[\mu_0]} = \frac{[M T^{-2} A^{-1}]}{[M L T^{-2} A^{-2}]}

Cancelling common powers of $M$ and $T$ ,

\frac{[B]}{[\mu_0]} = [L^{-1} A]

Therefore, the dimensions of $\frac{B}{\mu_0}$ are $L^{-1} A$ , so the correct option is B.

Common mistakes

Using an incorrect dimensional formula for magnetic field $B$ . The dimension of $B$ is $[M T^{-2} A^{-1}]$ , not one containing an extra power of $L$ . Start from the standard dimensional formula before dividing.
Forgetting that division of dimensions means subtracting exponents. In $\frac{[B]}{[\mu_0]}$ , powers of $M$ and $T$ cancel, and the power of $A$ becomes $-1-(-2)=1$ . Always subtract exponents carefully.
Confusing permeability $\mu_0$ with magnetic field intensity $H$ . From $B=\mu_0 H$ , the dimensions of $\frac{B}{\mu_0}$ are actually the dimensions of $H$ , which are $[A L^{-1}]$ . Use the relation correctly.

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