Units and Measurements JEE 2022: Magnetic Field Solution

What is the answer to the Units and Measurements JEE 2022 question?

The required answer is 4, the sum of all four exponents. This Units and Measurements JEE 2022 question is JEE Advanced 2022, Physics, Paper 2, Question 4, not a JEE Main question. It is a numerical-answer question; no answer options are supplied.

The task is to build magnetic-field dimensions using electric charge, electron mass, Planck’s constant and Coulomb’s constant, respectively:

e, m_{e}, h, k = \frac{1}{4 π ε_{0}}

The symbol for vacuum permittivity is: $ε_{0}$

For this dimensional identity, find the sum, not any individual exponent:

[B] = [e]^{α} [m_{e}]^{β} [h]^{γ} [k]^{δ}

$α + β + γ + δ = ?$ The question bank tags this as medium difficulty, with a 180-second practice target. Neither is an official exam instruction.

How do I derive the dimensions using mass, length, time and charge?

Use charge as a base dimension and derive magnetic-field dimensions from the magnetic force law. This follows the official solution and keeps every electromagnetic quantity in the same basis.

The base-dimension symbols mean:

M : mass, L : length, T : time, Q : charge

The capital charge symbol denotes a dimension, not the electric-charge quantity used as a building block. Thus:

[e] = Q, [m_{e}] = M

Planck’s constant has dimensions of energy multiplied by time:

[h] = (M L^{2} T^{- 2}) T = M L^{2} T^{- 1}

Start with Coulomb’s law to find the dimensions of Coulomb’s constant:

F = k \frac{q_{1} q_{2}}{r^{2}}

Rearrange dimensionally, keeping both charge factors:

[k] = \frac{[F] [r]^{2}}{[q]^{2}} = \frac{(M L T^{- 2}) L^{2}}{Q^{2}} = M L^{3} T^{- 2} Q^{- 2}

For magnetic force, use a charge moving perpendicular to the field: $F = q v B$

The general force magnitude includes a sine factor. That factor is dimensionless, so it does not change the dimensions:

[B] = \frac{[F]}{[q] [v]} = \frac{M L T^{- 2}}{Q L T^{- 1}} = M T^{- 1} Q^{- 1}

The length exponent is zero: the length factor in force cancels the length factor in velocity. Keep that zero when matching powers:

[B] = M^{1} L^{0} T^{- 1} Q^{- 1}

How do I substitute the dimensions and match all four powers?

Substitute each building block, collect powers of each base dimension, then match mass, length, time and charge separately. An absent base dimension has exponent zero, so the length equation must not be discarded.

First, insert all the dimensions:

M T^{- 1} Q^{- 1} = Q^{α} M^{β} (M L^{2} T^{- 1})^{γ} (M L^{3} T^{- 2} Q^{- 2})^{δ}

Now collect every contribution on the right:

M^{1} L^{0} T^{- 1} Q^{- 1} = M^{β + γ + δ} L^{2 γ + 3 δ} T^{- γ - 2 δ} Q^{α - 2 δ}

Matching powers gives:

Mass: & β + γ + δ = 1 Length: & 2 γ + 3 δ = 0 Time: & - γ - 2 δ = - 1 Charge: & α - 2 δ = - 1

The negative signs in the time and charge equations come directly from the input dimensions. Keep them attached to their terms rather than adjusting signs mentally after substitution.

How do I solve the exponents and independently verify 4?

Solve the time and length equations first, then use charge and mass. This follows the official elimination sequence and reduces the number of unknowns at each step. Check the result by substituting the final combination back into dimensions, not merely by adding the exponents again.

From time:

γ + 2 δ = 1 \Rightarrow γ = 1 - 2 δ

Substitute into length:

2 (1 - 2 δ) + 3 δ & = 0 2 - 4 δ + 3 δ & = 0 δ & = 2

Return to the expression for the Planck-constant exponent: $γ = 1 - 2 (2) = - 3$

Use charge next:

α - 2 (2) = - 1 \Rightarrow α = 3

Finally, use mass:

β + (- 3) + 2 = 1 \Rightarrow β = 2

The requested numerical entry is:

α + β + γ + δ = 3 + 2 - 3 + 2 = 4

For an independent check, form the combination specified by these exponents. The negative Planck-constant exponent puts its cube in the denominator:

[\frac{e^{3} m_{e}^{2} k^{2}}{h^{3}}] & = Q^{3} M^{2} \frac{M^{2} L^{6} T^{- 4} Q^{- 4}}{M^{3} L^{6} T^{- 3}} & = M T^{- 1} Q^{- 1}

The length powers cancel, and the remaining dimensions match magnetic field. Dimensional agreement is not a physical equality: it neither fixes a magnetic-field value nor determines a dimensionless numerical coefficient.

Why does using electric force give the wrong numerical answer?

Dropping velocity from the magnetic force law produces electric-field dimensions instead. The exponent equations can then be solved correctly while answering the wrong question. Since this PYQ has no options, the entry below is an illustrative wrong numerical answer, not an MCQ choice.

The incorrect starting point is:

F = q B \Rightarrow [B]_{wrong} = \frac{[F]}{[q]} = M L T^{- 2} Q^{- 1}

Force divided by charge has electric-field dimensions. Magnetic force requires the velocity factor: $F = q v B$

With the wrong target, the equations become:

2 γ + 3 δ & = 1 γ + 2 δ & = 2 β + γ + δ & = 1 α - 2 δ & = - 1

The mass and charge equations remain unchanged. Solving the altered time and length equations gives: $γ = 2 - 2 δ$

2 (2 - 2 δ) + 3 δ = 1 \Rightarrow 4 - δ = 1

δ = 3, γ = - 4, α = 5, β = 2

Consequently:

α + β + γ + δ = 5 + 2 - 4 + 3 = 6 \neq 4

The algebra has solved an electric-field problem. Verify the target dimensions from the defining force law before trusting a neatly solved exponent system.

Can I use the same method to find velocity and length combinations?

Keep the same mass–length–time–charge basis and change only the target dimensions. These are original practice questions, not additional verified PYQs. Attempt both before reading their worked checks.

Question 1: Find a combination of electric charge, electron mass, Planck’s constant and Coulomb’s constant with dimensions of velocity.
Question 2: Find a combination of the same constants with dimensions of length.

For both, use:

[e]^{α} [m_{e}]^{β} [h]^{γ} [k]^{δ}

Question 1 worked check: Match the powers for velocity:

β + γ + δ & = 0 2 γ + 3 δ & = 1 - γ - 2 δ & = - 1 α - 2 δ & = 0

Solve time and length first, then charge and mass:

γ = 1 - 2 δ, 2 - δ = 1

(α, β, γ, δ) = (2, 0, - 1, 1)

Verify directly:

[\frac{k e^{2}}{h}] = \frac{(M L^{3} T^{- 2} Q^{- 2}) Q^{2}}{M L^{2} T^{- 1}} = L T^{- 1}

Question 2 worked check: Match the powers for length:

β + γ + δ & = 0 2 γ + 3 δ & = 1 - γ - 2 δ & = 0 α - 2 δ & = 0

The same elimination order gives:

γ = - 2 δ, - δ = 1

(α, β, γ, δ) = (- 2, - 1, 2, - 1)

Verify the resulting combination:

[\frac{h^{2}}{m_{e} k e^{2}}] = \frac{M^{2} L^{4} T^{- 2}}{M^{2} L^{3} T^{- 2}} = L

If either check fails, compare the four base-dimension powers separately. Find the first mismatch before redoing the elimination.

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Frequently asked questions

Is the 2022 magnetic-field dimensional analysis question from JEE Main or Advanced?

It is from JEE Advanced 2022, Physics, Paper 2, Question 4, not JEE Main. It is a numerical-answer question with no answer options.

What is the answer to the Units and Measurements JEE 2022 magnetic-field question?

The required answer is 4, the sum of the four exponents. The exponents of electric charge, electron mass, Planck’s constant and Coulomb’s constant are 3, 2, -3 and 2, respectively.

What are the dimensions of magnetic field using charge as a base dimension?

Magnetic field has dimensions M T^-1 Q^-1, where M denotes mass, T time and Q charge. Derive this from F = qvB for motion perpendicular to the field by dividing force dimensions by charge and velocity dimensions. The length exponent is zero and must still be included when matching powers.

Why do I get 6 instead of 4 in the JEE 2022 magnetic-field question?

Using F = qB instead of F = qvB gives force-per-charge dimensions, which belong to electric field rather than magnetic field. Matching those incorrect target dimensions gives exponents 5, 2, -4 and 3, whose sum is 6. Retain the velocity factor to obtain the correct sum of 4.