Respuesta :
Answer:
(a). The rotational inertia is [tex]5.72\times10^{39}\ kg m^2[/tex]
(b). The magnitude of the magnetic torque is [tex]3.20\times10^{35}\ N-m[/tex]
Explanation:
Given that,
Mass of neutron [tex]M_{n}= 13M_{s}[/tex]
Density of neutron [tex]\rho=4.8\times10^{17}\ kg/m^3[/tex]
(a). We need to calculate the rotational inertia
Using formula of rotational inertia for sphere
[tex]I=\dfrac{2}{5}MR^2[/tex]...(I)
We know that,
[tex]\rho=\dfrac{M}{V}[/tex]
Put the value of volume
[tex]\rho=\dfrac{3M_{n}}{4\pi R^3}[/tex]
[tex]R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}[/tex]
Put the value of R in equation (I)
[tex]I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}[/tex]
Put the value into the formula
[tex]I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}[/tex]
[tex]I=5.72\times10^{39}\ kg m^2[/tex]
The rotational inertia is [tex]5.72\times10^{39}\ kg m^2[/tex].
(b). We need to calculate the magnitude of the magnetic torque
Using formula of torque
[tex]\tau=I\times \alpha[/tex]
Put the value into the formula
[tex]\tau=5.72\times10^{39}\times5.6\times10^{-5}[/tex]
[tex]\tau=3.20\times10^{35}\ N-m[/tex]
The magnitude of the magnetic torque is [tex]3.20\times10^{35}\ N-m[/tex]
Hence, (a). The rotational inertia is [tex]5.72\times10^{39}\ kg m^2[/tex]
(b). The magnitude of the magnetic torque is [tex]3.20\times10^{35}\ N-m[/tex]
