In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, re

Q: Why does $\sqrt{\mu_0/\epsilon_0}$ have the dimensions of resistance?

In electromagnetism, $\sqrt{\mu_0/\epsilon_0}$ is known as the impedance of free space. Impedance is a generalized form of resistance, so it shares the exact same dimensional formula: $[M L^2 T^{-3} A^{-2}]$.

Q: Can I just substitute 5 directly into the dimensional formula?

Yes, effectively! Since the new units are all 5 times the SI units, the conversion factor is simply $(1/5)$ raised to the respective powers of each dimension. But writing out $n_1 u_1 = n_2 u_2$ prevents silly mistakes.

Q: What if the new system had different scaling factors, like 2 kg and 5 m?

The process remains exactly the same. You would substitute $M_2 = 2\text{ kg}$ and $L_2 = 5\text{ m}$ into the ratio $(M_1/M_2)^1 (L_1/L_2)^2$, resulting in $(1/2)^1 (1/5)^2$.

Q: Why do we add the exponents when simplifying?

Because all the bases in the ratio are the same (1/5). According to the laws of exponents, when you multiply terms with the same base, you add their powers: $x^a \cdot x^b = x^{a+b}$.

Q: Is there a quicker way to find the dimensions of $\mu_0$ and $\epsilon_0$?

Yes! Use familiar formulas. For $\mu_0$, use the magnetic force per unit length: $F/L = (\mu_0 I^2)/(2\pi d)$. For $\epsilon_0$, use Coulomb's law: $F = (1/4\pi\epsilon_0)(q^2/r^2)$.

Question

In a new system of units, the units of mass, length, time and current are $5	ext{ kg}$, $5	ext{ m}$, $5	ext{ s}$ and $5	ext{ A}$, respectively. If $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{\mu_0 / \epsilon_0}$, is :

Sigma Percentile · Accepted Answer

### The Magic of Dimensional Analysis

Imagine you are an explorer who just landed on an alien planet. On this planet, the inhabitants don't use kilograms, meters, or seconds. Instead, their standard unit of mass is exactly $5\ \mathrm{kg}$, their unit of length is $5\ \mathrm{m}$, and so on. If you wanted to communicate the value of a physical constant to them, how would you translate it?

This is exactly what this problem asks us to do! We need to translate the magnitude of one SI unit of the quantity $\sqrt{\frac{\mu_0}{\epsilon_0}}$ into this new "alien" system of units. Don't panic; dimensional analysis is the universal translator we need.

### Decoding the Physical Quantity

Before we can translate anything, we need to know the "DNA" of the quantity we are dealing with. What are the dimensions of $\sqrt{\frac{\mu_0}{\epsilon_0}}$?

If you've studied electromagnetism, you might recognize this expression as the **impedance of free space** (often denoted as $Z_0 \approx 377\ \Omega$). Because it is an impedance, it shares the exact same dimensional formula as electrical resistance!

Let's quickly recall the dimensions of resistance $R$:
$$R = \frac{V}{I} = \frac{W/q}{I} = \frac{	ext{Work}}{	ext{Charge} 	imes 	ext{Current}}$$

Substituting the fundamental dimensions:
$$[R] = \frac{M L^2 T^{-2}}{A T \cdot A} = M^1 L^2 T^{-3} A^{-2}$$

So, the dimensional formula for our quantity is $[M^1 L^2 T^{-3} A^{-2}]$. This is the blueprint we will use for our conversion.

### The Universal Conversion Principle

The golden rule of unit conversion is that the actual physical amount of a quantity doesn't change just because you measure it differently. A stick is the same length whether you measure it in inches or centimeters. Mathematically, this is expressed as:
$$n_1 u_1 = n_2 u_2$$

Here, $n_1$ and $n_2$ are the numerical values, and $u_1$ and $u_2$ are the sizes of the units in the respective systems.

We are given that in the SI system, the magnitude is $1$ (so $n_1 = 1$). We need to find $n_2$ in the new system. Let's set up our equation using the dimensional blueprint:
$$1 \cdot \left[ M_1^1 L_1^2 T_1^{-3} A_1^{-2} ight] = n_2 \cdot \left[ M_2^1 L_2^2 T_2^{-3} A_2^{-2} ight]$$

### The Final Calculation

Now, we isolate our target, $n_2$, by grouping the corresponding base units together:
$$n_2 = 1 \cdot \left(\frac{M_1}{M_2}ight)^1 \left(\frac{L_1}{L_2}ight)^2 \left(\frac{T_1}{T_2}ight)^{-3} \left(\frac{A_1}{A_2}ight)^{-2}$$

We know the relationship between the old (SI) units and the new units. The new units are all exactly $5$ times the SI units! So, $M_2 = 5 M_1$, $L_2 = 5 L_1$, and so forth. Substituting these ratios gives:
$$n_2 = \left(\frac{1}{5}ight)^1 \left(\frac{1}{5}ight)^2 \left(\frac{1}{5}ight)^{-3} \left(\frac{1}{5}ight)^{-2}$$

Notice how beautifully the units cancel out, leaving us with a pure arithmetic problem. Since all the bases are the same ($\frac{1}{5}$), we can simply add the exponents:
$$	ext{Sum of exponents} = 1 + 2 - 3 - 2 = -2$$

So, our equation simplifies to:
$$n_2 = \left(\frac{1}{5}ight)^{-2}$$

A negative exponent means we take the reciprocal, so:
$$n_2 = 5^2 = 25$$

**And there we have it!** One SI unit of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to exactly $25$ units in this new system. By trusting the dimensional formula and the principle of conversion, what looked like a complex electromagnetic problem melted into simple arithmetic.

Enter Numerical Value:

Visualized Solution

$[\frac{μ _{0}}{ϵ _{0}}]$

$n_{1} u_{1} = n_{2} u_{2}$

$1 \cdot u_{1} = n_{2} \cdot u_{2}$

$n_{2} = \frac{u _{1}}{u _{2}}$

$n_{2} = \prod (\frac{1}{5})^{x}$

$x^{a} \cdot x^{b} = x^{a + b}$

$n_{2} = 25$

$M_{2} = 10 kg$

The Sigma Insight: Dimensional Analysis

The Magic of Dimensional Analysis

Decoding the Physical Quantity

The Universal Conversion Principle

The Final Calculation

Similar Questions

Let $[ε_{0}]$ denote the dimensional formula of the permittivity of the vacuum and $[μ_{0}]$ that of the permeability of the vacuum. If $M = mass$ , $L = length$ , $T = time$ and $I = electric current$ .

Comprehension Passage

The relation between $[ε_{0}]$ and $[μ_{0}]$ is

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The relation between $[E]$ and $[B]$ is

The dimensions of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ : permittivity of free space; $E$ : electric field) is

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In terms of potential difference $V$ , electric current $I$ , permittivity $ε_{0}$ , permeability $μ_{0}$ and speed of light $c$ , the dimensionally correct equations is/are

A quantity $X$ is given by $ε_{0} L \frac{Δ V}{Δ t}$ , where $ε_{0}$ is the permittivity of free space, $L$ is a length, $Δ V$ is a potential difference and $Δ t$ is a time interval. The dimensional formula for $X$ is the same as that of

Write the dimensions of the following in terms of mass, time, length and charge. (a) Magnetic flux (b) Rigidity modulus

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [ϵ0​μ0​​​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​u1​=n2​u2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1⋅u1​=n2​⋅u2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​=u2​u1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​=∏(51​)x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xa⋅xb=xa+b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​=25

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2​=10 kg

The Sigma Insight: Dimensional Analysis

The Magic of Dimensional Analysis

Decoding the Physical Quantity

The Universal Conversion Principle

The Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The relation between [ε0​] and [μ0​] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [ε0​] denotes the dimensional formula of the permittivity of vacuum. If M=mass, L=length, T=time and A=electric current, then

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The relation between [E] and [B] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The dimensions of 21​ε0​E2 (ε0​ : permittivity of free space; E : electric field) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The dimension of 2μ0​B2​, where B is magnetic field and μ0​ is the magnetic permeability of vacuum, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In terms of potential difference V, electric current I, permittivity ε0​, permeability μ0​ and speed of light c, the dimensionally correct equations is/are

Write the dimensions of the following in terms of mass, time, length and charge. (a) Magnetic flux (b) Rigidity modulus

$[\frac{μ _{0}}{ϵ _{0}}]$

$n_{1} u_{1} = n_{2} u_{2}$

$1 \cdot u_{1} = n_{2} \cdot u_{2}$

$n_{2} = \frac{u _{1}}{u _{2}}$

$n_{2} = \prod (\frac{1}{5})^{x}$

$x^{a} \cdot x^{b} = x^{a + b}$

$n_{2} = 25$

$M_{2} = 10 kg$

The relation between $[ε_{0}]$ and $[μ_{0}]$ is

Let $[ε_{0}]$ denotes the dimensional formula of the permittivity of vacuum. If $M = mass$ , $L = length$ , $T = time$ and $A = electric current$ , then

The relation between $[E]$ and $[B]$ is

The dimensions of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ : permittivity of free space; $E$ : electric field) is

The dimension of $\frac{B ^{2}}{2 μ _{0}}$ , where $B$ is magnetic field and $μ_{0}$ is the magnetic permeability of vacuum, is

In terms of potential difference $V$ , electric current $I$ , permittivity $ε_{0}$ , permeability $μ_{0}$ and speed of light $c$ , the dimensionally correct equations is/are