Question

A neutron star with magnetic moment of magnitude $m$ is spinning with angular velocity $\omega$ about its magnetic axis. The electromagnetic power $P$ radiated by it is given by ${\mu }_0^x{m}^y{\omega }^z{c}^u$, where ${\mu }_0$ and $c$ are the permeability and speed of light in free space, respectively. Then,

• A. $x=1,y=2,z=4$ and $u=-3$
• B. $x=1,y=2,z$= 4 and $u=3$
• C. $x=-1y=2,z=4$ and $u=-3$
• D. $x=-1,y=2,z=4$ and u = 34

Answer 1

Answer: (A)

Given, power radiated $P$ is
$P={\mu }_0^x{m}^y{\omega }^z{c}^u$
Substituting dimensions of different physical quantities involved, we have
$\left[M{L}^2{T}^{-3}\right]={\left[ML{T}^{-2}{A}^{-2}\right]}^x{\left[L^{2}A\right]}^y{\left[T^{-1}\right]}^z{\left[L{T}^{-1}\right]}^u$
Equating powers of fundamental quantities, we have
$x=\mathrm{1...}\left(i\right)$
$x+2y+u=2\left(ii\right)$
$-2x-z-u=-\mathrm{3...}\left(iii\right)$
$-2x+y=\mathrm{0...}\left(iv\right)$
From Eq. (i), putting the value of $x$ in Eq. (iv), we get
$\Rightarrow -2\times 1+y=0$
$\Rightarrow y=\mathrm{2...}\left(v\right)$
Now, from Eqs. (i) and (v),
putting the values of $x$ and $y$ in Eq. (ii), we get
$\Rightarrow 1+2\times 2+u=2$
$\Rightarrow u=-\mathrm{3...}\left(vi\right)$
Now, again from Eqs. (i) and (vi),
putting the values of $x$ and $u$ in Eq. (iii), we get
$\Rightarrow -2\times 1-z+3=-3$
$\Rightarrow z=4$
So $x=1,y=2,z=4,u=-3,$