Units and Measurements JEE 2021: B and D Explained

What is the answer to the Units and Measurements JEE 2021 question?

B and D are correct in this Units and Measurements JEE 2021 problem from JEE Advanced 2021, Paper 2, Physics. More than one option can be correct. The required dimensions are: $[S] = M T^{- 3}$

An electric field and a magnetic field define a vector quantity through their cross product:

\vec{S} = \frac{\vec{E} \times \vec{B}}{μ_{0}}

The denominator denotes the permeability of free space. Determine which listed expressions share the dimensions of this quantity:

How do you derive electric and magnetic field dimensions from force?

Use force per charge for the electric field and force on a current-carrying wire for the magnetic field. These are the official solution’s starting relations; neither field’s dimensions need to be memorised.

The dimensional symbols represent mass, length, time and electric current. The current symbol below is unrelated to option A.

M : mass, L : length, T : time, A : electric current

Start with force and charge, using charge as current multiplied by time:

[F] = M L T^{- 2}, [q] = A T

Electric field is force per unit charge:

[E] = \frac{[F]}{[q]} = \frac{M L T^{- 2}}{A T} = M L T^{- 3} A^{- 1}

For a straight wire perpendicular to a magnetic field, force is field strength multiplied by current and wire length: $F = B I L$

For another orientation, the force includes a sine factor. That factor is dimensionless, so it does not change the calculation.

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

How do you derive the dimensions of permeability of free space?

Use the official solution’s relation for the magnetic field near a long straight current-carrying wire. Rearranging it isolates permeability using the magnetic-field dimensions just derived.

B = \frac{μ_{0} I}{2 π r}

The radial distance has dimensions of length, while the numerical factor is dimensionless:

[r] = L, [2 π] = 1

Rearranging dimensionally gives:

[μ_{0}] = \frac{[B] [r]}{[I]}

Substitute every factor, including the current in the denominator:

[μ_{0}] = \frac{M T^{- 2} A^{- 1} \times L}{A} = M L T^{- 2} A^{- 2}

A constant is not automatically dimensionless. Permeability retains these dimensions even though it is a constant.

How do the dimensions combine to give the final result?

Multiply the field dimensions and divide by the permeability dimensions. The cross product determines direction and introduces a dimensionless angular factor, so dimensionally it gives the product of the field dimensions. $[E \times B] = [E] [B]$

Write the full substitution before cancelling:

[S] = \frac{[E] [B]}{[μ_{0}]} = \frac{(M L T^{- 3} A^{- 1}) (M T^{- 2} A^{- 1})}{M L T^{- 2} A^{- 2}}

First collect the numerator:

[E] [B] = M^{2} L T^{- 5} A^{- 2}

Then subtract each denominator exponent from its corresponding numerator exponent:

[S] = M^{2 - 1} L^{1 - 1} T^{- 5 + 2} A^{- 2 + 2} = M T^{- 3}

Both length and current cancel completely. Subtracting the negative time exponent in the denominator produces the positive two in the time calculation.

The equivalent SI unit is: $k g s^{- 3}$

Why do options B and D match, but A and C do not?

Only force divided by length and time, and power divided by area, reduce to the required dimensions. Check all four expressions independently rather than stopping at the first match.

A: Energy divided by the product of charge and current. It does not match: length and current remain.

\frac{M L^{2} T^{- 2}}{(A T) A} = M L^{2} T^{- 3} A^{- 2}

B: Force divided by the product of length and time. It matches.

\frac{M L T^{- 2}}{L T} = M T^{- 3}

C: Energy divided by volume. It does not match: these are energy-density dimensions.

\frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}

D: Power divided by area. First obtain power from energy per time, then divide by area. It matches.

[Power] = \frac{[Energy]}{[Time]} = \frac{M L^{2} T^{- 2}}{T} = M L^{2} T^{- 3}

\frac{M L^{2} T^{- 3}}{L^{2}} = M T^{- 3}

As a separate SI unit check:

W m^{- 2} = \frac{k g m^{2} s^{- 3}}{m^{2}} = k g s^{- 3}

For a numerical check, take power spread uniformly over an area:

P = 12 W, A = 3 m^{2}

\frac{P}{A} = \frac{12}{3} = 4 W m^{- 2} = 4 k g s^{- 3}

The unit check supports the derivation; it does not replace it. Select B and D. One matching option is not permission to stop in a multiple-correct question.

Why is energy density the wrong interpretation for option C?

Energy density lacks a factor with the dimensions of speed. Option C is energy per volume; option D is energy per time per area.

A possible faulty method is to recall that the following expression has energy-density dimensions, then assign those dimensions to the given quantity without checking its different numerator:

\frac{B^{2}}{μ_{0}}

For nonzero field magnitudes, the missing factor is visible in the scalar product:

\frac{E B}{μ_{0}} = \frac{B^{2}}{μ_{0}} \frac{E}{B}

The extra ratio is not dimensionless. It has the dimensions of velocity:

\frac{[E]}{[B]} = \frac{M L T^{- 3} A^{- 1}}{M T^{- 2} A^{- 1}} = L T^{- 1}

Multiplying energy-density dimensions by velocity dimensions gives:

(M L^{- 1} T^{- 2}) (L T^{- 1}) = M T^{- 3}

This distinguishes energy density from energy flux. The dimensional comparison needs neither an electromagnetic-wave assumption nor a relation involving the speed of light.

How can I practise this dimensional-analysis method?

Reuse the field and permeability dimensions to distinguish pressure from power per area, then test what dimensional agreement can prove. These are original practice questions from the same chapter, not additional verified PYQs.

Determine the dimensions of the following quantity. Does it match pressure or power per area?

X = \frac{B^{2}}{2 μ_{0}}

The factor two is dimensionless. Substituting the derived dimensions gives:

[X] = \frac{(M T^{- 2} A^{- 1})^{2}}{M L T^{- 2} A^{- 2}} = M L^{- 1} T^{- 2}

Pressure is force per area:

[Pressure] = \frac{M L T^{- 2}}{L^{2}} = M L^{- 1} T^{- 2}

The result matches pressure, not power per area.

For nonzero field magnitudes, determine the dimensions of electric field divided by magnetic field. Does this alone prove that the ratio equals the speed of light?

[\frac{E}{B}] = \frac{M L T^{- 3} A^{- 1}}{M T^{- 2} A^{- 1}} = L T^{- 1}

The ratio has the dimensions of speed. Dimensional agreement alone establishes neither a numerical value nor a physical equality.

Cover the worked solution and rebuild the field and permeability dimensions from their defining equations. Then check all four original options without stopping at B.

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Frequently asked questions

What is the answer to the Units and Measurements JEE 2021 question?

B and D are correct in this JEE Advanced 2021 Paper 2 Physics question. The quantity S = (E × B)/μ₀ has dimensions MT⁻³. Both force divided by length and time, and power divided by area, have these dimensions.

How do I derive the dimensions of permeability of free space?

Start with the magnetic field near a long straight wire: B = μ₀I/(2πr). Rearranging dimensionally gives [μ₀] = [B][r]/[I] = MLT⁻²A⁻², where A represents electric current. Permeability is a constant, but it is not dimensionless.

Why is energy density not the correct answer?

Energy density has dimensions ML⁻¹T⁻², whereas the given quantity has dimensions MT⁻³. The ratio E/B has dimensions of speed, so replacing EB/μ₀ with B²/μ₀ loses a dimensional factor. The given quantity has energy-flux dimensions, not energy-density dimensions.

Does dimensional analysis prove that E/B equals the speed of light?

No. For nonzero field magnitudes, E/B has dimensions LT⁻¹, which are the dimensions of speed. Dimensional agreement alone does not establish a numerical value or prove a physical equality.

Units and Measurements JEE 2021: B and D Explained

What is the answer to the Units and Measurements JEE 2021 question?

How do you derive electric and magnetic field dimensions from force?

How do you derive the dimensions of permeability of free space?

How do the dimensions combine to give the final result?

Why do options B and D match, but A and C do not?

Why is energy density the wrong interpretation for option C?

How can I practise this dimensional-analysis method?

Frequently asked questions

What is the answer to the Units and Measurements JEE 2021 question?

How do I derive the dimensions of permeability of free space?

Why is energy density not the correct answer?

Does dimensional analysis prove that E/B equals the speed of light?

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