Match List-I with List-II.
List-I
A. Magnetic induction
B. Magnetic flux
C. Magnetic permeability
D.

Q: Why did we use $F = qvB$ for magnetic induction? Can we use another formula?

Yes, you can use any valid formula containing $B$, such as $F = I L B$. The dimensional formula will always come out the same. $F = qvB$ is just one of the most fundamental and easy-to-remember equations.

Q: What is the difference between magnetic field $B$ and magnetic flux $\phi$?

Magnetic field $B$ (or magnetic induction) is the strength of the field at a point. Magnetic flux $\phi$ is the total magnetic field passing through a given area. Dimensionally, flux is simply the magnetic field multiplied by area ($[L^2]$).

Q: In the formula $B = \mu_0 n I$, what is the dimension of $n$?

Here, $n$ represents the number of turns per unit length of the solenoid. Since it is a count (dimensionless) divided by length, its dimension is $[L^{-1}]$.

Q: What exactly is Magnetisation?

Magnetisation represents the density of permanent or induced magnetic dipole moments in a magnetic material. It is calculated as the total magnetic moment divided by the volume, hence its formula $I = \frac{M}{V}$.

Q: What should I do in an exam if I spot a typo in the options like this one?

If you are confident in your derivation, look for the option that matches all the other correct pairs. Mark that option. Often, exam boards will either award bonus marks to everyone or to those who marked the closest logical answer.

Question

Match List-I with List-II.

$\begin{array}{ll}
	ext{List-I} & 	ext{List-II} \
	ext{A. Magnetic induction} & 	ext{1. } [ML^2T^{-2}A^{-1}] \
	ext{B. Magnetic flux} & 	ext{2. } [ML^{-1}A] \
	ext{C. Magnetic permeability} & 	ext{3. } [MT^{-2}A^{-1}] \
	ext{D. Magnetisation} & 	ext{4. } [MLT^{-2}A^{-2}]
\end{array}$

Choose the most appropriate answer from the options given below.

Options:

(A) A $ightarrow$ 2, B $ightarrow$ 4, C $ightarrow$ 1, D $ightarrow$ 3

(B) A $ightarrow$ 2, B $ightarrow$ 1, C $ightarrow$ 4, D $ightarrow$ 3

(C) A $ightarrow$ 3, B $ightarrow$ 2, C $ightarrow$ 4, D $ightarrow$ 1

(D) A $ightarrow$ 3, B $ightarrow$ 1, C $ightarrow$ 4, D $ightarrow$ 2

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Dimensional analysis is one of the most reliable tools in a physicist's arsenal. It allows us to verify equations, convert units, and even guess the form of unknown relationships.

In this problem, we are tasked with matching four fundamental magnetic quantities with their dimensional formulas.

It seems like a straightforward exercise, but as we will discover, even official exam papers can sometimes contain a hidden trap!

### Analyzing Magnetic Induction

Let's start with the first quantity: **Magnetic Induction**, denoted by $B$.

To find its dimensions, we need a formula that connects $B$ to quantities whose dimensions we already know. The most fundamental equation is the Lorentz force on a moving charge:
$$F = qvB$$

Rearranging this to solve for $B$, we get:
$$B = \frac{F}{qv}$$

Now, we substitute the known dimensional formulas. Force $F$ is $[MLT^{-2}]$, charge $q$ is $[AT]$, and velocity $v$ is $[LT^{-1}]$.

$$[B] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]}$$

Notice how the length dimension $L$ cancels out completely.

$$[B] = [MT^{-2}A^{-1}]$$

This perfectly matches option 3 in List-II.

### Deriving Magnetic Flux

Next up is **Magnetic Flux**, denoted by $\phi$.

Flux is simply the measure of the total magnetic field passing through a given area. Mathematically, it is the dot product of the magnetic field vector and the area vector:
$$\phi = B \cdot A$$

Since we just calculated the dimension of $B$, this step is a breeze. The dimension of area $A$ is simply $[L^2]$.

$$[\phi] = [MT^{-2}A^{-1}] 	imes [L^2]$$

Combining these, we get:
$$[\phi] = [ML^2T^{-2}A^{-1}]$$

Looking at List-II, this is an exact match for option 1.

### Uncovering Magnetic Permeability

Our third quantity is **Magnetic Permeability**, $\mu_0$.

To find its dimensions, imagine a long, tightly wound solenoid. The magnetic field inside an ideal solenoid is given by:
$$B = \mu_0 n I$$

Here, $n$ is the number of turns per unit length, and $I$ is the current. We can rearrange this to solve for $\mu_0$:
$$\mu_0 = \frac{B}{nI}$$

We already know $[B]$. The dimension of $n$ is $[L^{-1}]$ because it is a count divided by length. The dimension of current $I$ is simply $[A]$.

$$[\mu_0] = \frac{[MT^{-2}A^{-1}]}{[L^{-1}][A]}$$

When we bring the terms from the denominator to the numerator, the signs of their exponents flip.

$$[\mu_0] = [MLT^{-2}A^{-2}]$$

This matches option 4 perfectly.

### The Trap of Magnetisation

Finally, we arrive at **Magnetisation**, denoted by $I$ or $M$.

Magnetisation is defined as the magnetic dipole moment acquired per unit volume of a material. 
$$I = \frac{M}{V}$$

The magnetic moment $M$ is the product of current and area, so its dimension is $[L^2A]$. The volume $V$ has a dimension of $[L^3]$.

$$[I] = \frac{[L^2A]}{[L^3]}$$

Simplifying this gives:
$$[I] = [L^{-1}A]$$

Now, let's look at option 2 in List-II. It says $[ML^{-1}A]$.

**There is an extra $M$!**

This is a typographical error in the official exam paper. The mass dimension should not be there.

### Final Conclusion

Because of this typo, technically none of the given options are 100% correct.

However, in competitive exams like JEE, you must choose the *most appropriate* answer. If we assume that option 2 was intended to be $[L^{-1}A]$, then our matching is:
- A $ightarrow$ 3
- B $ightarrow$ 1
- C $ightarrow$ 4
- D $ightarrow$ 2

This corresponds exactly to option (d). Always trust your derivations, and don't let a small typo derail your confidence!

Select Answer:

Visualized Solution

$Matrix Match Setup$

$Magnetic Induction (B)$

$Magnetic Flux (ϕ)$

$Magnetic Permeability (μ_{0})$

$Magnetisation (I)$

$Conclusion$

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Analyzing Magnetic Induction

Deriving Magnetic Flux

Uncovering Magnetic Permeability

The Trap of Magnetisation

Final Conclusion

Similar Questions

Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

Match the physical quantities given in Column I with dimensions expressed in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and charge ( $Q$ ) given in Column II and write the correct answer against the matched quantity in a tabular form in your answer book.

List-I

List-II

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

Column I gives three physical quantities. Select the appropriate units for the choices given in Column II. Some of the physical quantities may have more than one choice.

List-I

List-II

The dimensions of magnetic field in $M, L, T$ and $C$ (coulomb) is given as

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

Give the MKS units for each of the following quantities. (a) Young's modulus (b) Magnetic induction (c) Power of a lens

Comprehension Passage

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

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Matrix Match Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Induction (B)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Flux (ϕ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Permeability (μ0​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetisation (I)

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

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Analyzing Magnetic Induction

Deriving Magnetic Flux

Uncovering Magnetic Permeability

The Trap of Magnetisation

Final Conclusion

Similar Questions

Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

List-I

List-II

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

Column I gives three physical quantities. Select the appropriate units for the choices given in Column II. Some of the physical quantities may have more than one choice.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The dimensions of magnetic field in M,L,T and C (coulomb) is given as

.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

Give the MKS units for each of the following quantities. (a) Young's modulus (b) Magnetic induction (c) Power of a lens

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

$Matrix Match Setup$

$Magnetic Induction (B)$

$Magnetic Flux (ϕ)$

$Magnetic Permeability (μ_{0})$

$Magnetisation (I)$

$Conclusion$

The dimensions of magnetic field in $M, L, T$ and $C$ (coulomb) is given as

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

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