A physical quantity \overrightarrow S is defined as \overrightarrow S = (\overrightarrow E \times \overrightarrow B )/\mu _0 , where \overrightarrow E is electric field, \overrightarrow B is magnetic field and \mu 0 is the permeability of free space. The dimensions of \overrightarrow S are the same as the dimensions of which of the following quantity(ies)?

Force \over Length \times Time, Power \over Area

Energy \over Ch\arg e \times Current

Units and Measurements PYQ — JEE Advanced 2021 · Shift 2 · Q42 with Solution

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Correct answer: B, D

Given quantity

We need the dimensions of

\vec S = \frac{\vec E \times \vec B}{\mu_0}

Since cross product does not affect dimensions,

[S] = \frac{[E][B]}{[\mu_0]}

Use physical meaning of $\vec S$

This is the magnitude form of the Poynting vector, whose SI unit is energy flow per unit area per unit time, i.e.

\text{W m}^{-2}

So dimensionally,

[S] = \frac{\text{Power}}{\text{Area}}

Hence Option D is correct.

But let us also verify by dimensions directly and compare with other options.

Dimensional formula of $S$

Using known dimensions:

Electric field,

[E] = \frac{\text{Force}}{\text{Charge}} = \frac{MLT^{-2}}{IT} = MLT^{-3}I^{-1}

Magnetic field from Lorentz force $F=qvB$ ,

[B] = \frac{F}{qv} = \frac{MLT^{-2}}{(IT)(LT^{-1})} = MT^{-2}I^{-1}

Permeability of free space,

[\mu_0] = ML T^{-2} I^{-2}

(For example from $F/L = \mu_0 I^2/(2\pi r)$ .)

Therefore,

[S] = \frac{(MLT^{-3}I^{-1})(MT^{-2}I^{-1})}{MLT^{-2}I^{-2}}

[S] = M L^0 T^{-3}

So,

[S] = MT^{-3}

Check each option

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Dimensions:

\frac{ML^2T^{-2}}{(IT)(I)} = ML^2T^{-3}I^{-2}

This is not equal to $MT^{-3}$ .

❌ A is incorrect.

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Dimensions:

\frac{MLT^{-2}}{L\,T} = MT^{-3}

This matches $[S]$ .

✅ B is correct.

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Dimensions:

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

This is not equal to $MT^{-3}$ .

❌ C is incorrect.

Option D: $\dfrac{\text{Power}}{\text{Area}}$

Dimensions:

\frac{ML^2T^{-3}}{L^2} = MT^{-3}

This matches $[S]$ .

✅ D is correct.

Final answer

The quantities having the same dimensions as $\vec S$ are:

\boxed{B,\ D}

Comparison with stored correct answer

Stored correct answer: B, D

Our derived answer: B, D

So the answer agrees with the stored correct answer.

More from Units and Measurements

View written solutionFree

Correct answer: B, D

Given quantity

We need the dimensions of

\vec S = \frac{\vec E \times \vec B}{\mu_0}

Since cross product does not affect dimensions,

[S] = \frac{[E][B]}{[\mu_0]}

Use physical meaning of $\vec S$

This is the magnitude form of the Poynting vector, whose SI unit is energy flow per unit area per unit time, i.e.

\text{W m}^{-2}

So dimensionally,

[S] = \frac{\text{Power}}{\text{Area}}

Hence Option D is correct.

But let us also verify by dimensions directly and compare with other options.

Dimensional formula of $S$

Using known dimensions:

Electric field,

[E] = \frac{\text{Force}}{\text{Charge}} = \frac{MLT^{-2}}{IT} = MLT^{-3}I^{-1}

Magnetic field from Lorentz force $F=qvB$ ,

[B] = \frac{F}{qv} = \frac{MLT^{-2}}{(IT)(LT^{-1})} = MT^{-2}I^{-1}

Permeability of free space,

[\mu_0] = ML T^{-2} I^{-2}

(For example from $F/L = \mu_0 I^2/(2\pi r)$ .)

Therefore,

[S] = \frac{(MLT^{-3}I^{-1})(MT^{-2}I^{-1})}{MLT^{-2}I^{-2}}

[S] = M L^0 T^{-3}

So,

[S] = MT^{-3}

Check each option

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Dimensions:

\frac{ML^2T^{-2}}{(IT)(I)} = ML^2T^{-3}I^{-2}

This is not equal to $MT^{-3}$ .

❌ A is incorrect.

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Dimensions:

\frac{MLT^{-2}}{L\,T} = MT^{-3}

This matches $[S]$ .

✅ B is correct.

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Dimensions:

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

This is not equal to $MT^{-3}$ .

❌ C is incorrect.

Option D: $\dfrac{\text{Power}}{\text{Area}}$

Dimensions:

\frac{ML^2T^{-3}}{L^2} = MT^{-3}

This matches $[S]$ .

✅ D is correct.

Final answer

The quantities having the same dimensions as $\vec S$ are:

\boxed{B,\ D}

Comparison with stored correct answer

Stored correct answer: B, D

Our derived answer: B, D

So the answer agrees with the stored correct answer.

Units and Measurements question

Units and Measurements question

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Option D: $\dfrac{\text{Power}}{\text{Area}}$

More from Units and Measurements

Units and Measurements question

Units and Measurements question

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Option D: $\dfrac{\text{Power}}{\text{Area}}$

More from Units and Measurements

Option A: EnergyCharge×Current\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}Charge×CurrentEnergy​

Option B: ForceLength×Time\dfrac{\text{Force}}{\text{Length} \times \text{Time}}Length×TimeForce​

Option C: EnergyVolume\dfrac{\text{Energy}}{\text{Volume}}VolumeEnergy​

Option D: PowerArea\dfrac{\text{Power}}{\text{Area}}AreaPower​

More from Units and Measurements

Option A: EnergyCharge×Current\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}Charge×CurrentEnergy​

Option B: ForceLength×Time\dfrac{\text{Force}}{\text{Length} \times \text{Time}}Length×TimeForce​

Option C: EnergyVolume\dfrac{\text{Energy}}{\text{Volume}}VolumeEnergy​

Option D: PowerArea\dfrac{\text{Power}}{\text{Area}}AreaPower​

More from Units and Measurements

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Option D: $\dfrac{\text{Power}}{\text{Area}}$

Option A: $\dfrac{\text{Energy}}{\text{Charge} \times \text{Current}}$

Option B: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$

Option C: $\dfrac{\text{Energy}}{\text{Volume}}$

Option D: $\dfrac{\text{Power}}{\text{Area}}$