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

Answer 1

Answer: (A)

Given, power radiated $P$ is
$P=\mu^{x}_{0}\,m^{y}\,\omega^{z}\,c^{u}$
Substituting dimensions of different physical quantities involved, we have
$\left[ML^{2}T^{-3}\right]=\left[MLT^{-2}A^{-2}\right]^{x}\left[L^{2 }A\right]^{y}\left[T^{-1}\right]^{z}\left[LT^{-1}\right]^{u}$
Equating powers of fundamental quantities, we have
$x=1 ...\left(i\right)$
$x+2y+u=2 \left(ii\right)$
$-2x-z-u=-3 ...\left(iii\right)$
$-2x+y=0 ...\left(iv\right)$
From Eq. (i), putting the value of $x$ in Eq. (iv), we get
$\Rightarrow -2\times1+y=0$
$\Rightarrow y=2 ...\left(v\right)$
Now, from Eqs. (i) and (v),
putting the values of $x$ and $y$ in Eq. (ii), we get
$\Rightarrow 1+2\times2+u=2$
$\Rightarrow u=-3 ...\left(vi\right)$
Now, again from Eqs. (i) and (vi),
putting the values of $x$ and $u$ in Eq. (iii), we get
$\Rightarrow -2\times1-z+3=-3 $
$\Rightarrow z=4$
So $x=1, y=2, z=4, u=-3,$