Question

Answer the question which is given in the attachment.

Answer 1

Here, $\mu_{\circ}$ refers to Absolute Permeability. It comes under the concept of Magnetism and Magnetic Fields.


There is one formula, known as Biot-Savart Law, which is:

$d\vec{B} = \frac{\mu_{\circ} I}{4\pi} \, \, \frac{\vec{dl} \times \vec{r}}{| \vec{r} |^3} \\ \\ \\ \implies \vec{B} = \frac{\mu_{\circ} I}{4\pi} \, \, \int \frac{\vec{dl} \times \vec{r}}{|\vec{r} |^3}$

Here, $B$ is Magnetic Field, $I$ is Current, $\vec{dl}$ is a position vector with direction along that of Current, and $\vec{r}$ is the position vector of the point where Magnetic Field $B$ is to be found. 

As you can yourself see, if we want to get answer from this formula, we will need dimensions of magnetic field B, which poses another challenge. It will require use of some other formula, like 

Force due to Magnetic Field B on a charge q, moving at a velocity v is given by:
$\vec{F} = q \, \, \vec{v} \times \vec{B}$

You can try it from here. 


Or We Will use another method:

___________________________________________


We will use the formula:

$\boxed{c=\frac{1}{\sqrt{\mu_{\circ} \varepsilon_{\circ}}}}$

We will need $\varepsilon_{\circ}$, which we will find from Coulomb's Law:


$F = \frac{1}{4\pi \varepsilon_{\circ}} \, \frac{q_1 q_2}{r^2} \\ \\ \\ \implies \varepsilon_{\circ} = \frac{1}{4\pi F} \, \frac{q_1 q_2}{r^2} \\ \\ \\ \implies [ \, \varepsilon_{\circ} \, ] = \frac{[ \, Q^2 \, ]}{[ \, F \, r^2 \, ]}$

Now, the unit of Force is $kg \, m/s^2$. So unit is $[\, M \, L \, T^{-2} \, ]$
Also, Charge is Q = Current $\times$ Time. 

So, dimensions of Charge are [ A T ]


Now,

$[ \, \varepsilon_{\circ} \, ] = \frac{[ \, Q^2 \, ]}{[ \, F \, r^2 \, ]} \\ \\ \\ \implies [ \, \varepsilon_{\circ} \, ] = \frac{[ \, A^2 \, T^2 \, ]}{[ \, M \, L \, T^{-2} \, ] \, \, [ \, L^2 \, ]} \\ \\ \\ \implies [ \, \varepsilon_{\circ} \, ] = [ \, M^{-1} \, L^{-3} \, T^4 \, A^2 \, ]$

Now, we can use the formula and find our answer.

$c = \frac{1}{\sqrt{\mu_{\circ}\varepsilon_{\circ}}}$

Here, c is speed of light. Its dimensions are $[\, L \, T^{-1} \, ]$ as its unit is metres per second. 

$\mu_{\circ} = \frac{1}{c^2 \varepsilon_{\circ}} \\ \\ \\ \implies [ \, \mu_{\circ} \, ] = \frac{1}{[ \, c^2 \varepsilon_{\circ} \, ]} \\ \\ \\ \implies [ \, \mu_{\circ} \, ] = \frac{1}{[\,L^2 \, T^{-2} \, ] \, \, [ \, M^{-1} \, L^{-3} \, T^4 \, A^2 \, ] } \\ \\ \\ \implies \boxed{[ \, \mu_{\circ} \, ] = [ \, M^1 \, L^1 \, T^{-2} \, A^{-2} \, ]}$


Thus, the dimensional formula for $\mu_{\circ}$ is Option (4) , which is $[ \, M^1 \, L^1 \, T^{-2} \, A^{-2} \, ]$




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