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