# Tritium & Deuterium. Coil Torque

#### Tritium and Deuterium

Please find  energy that is necessary to remove a neutron of tritium ($^3H$) to give a deuterium atom ($^2H$) and a neutron.

The rest energy for: neutron = 939.565 MeV, proton = 938.272 MeV, deuterium = 1875.613 MeV

$3H +energy = 2H +1n$

atomic masses are (all are taken from wiki)

$M(3H) =3.01604 uam =2817.143 MeV$ for tritium

$M(2H) = 2.0141 uam = 1881.27 MeV$   for deuterium

$M(1n) = 1.00866 uam = 942.142 MeV$

because

$1 uam =1.6605*10^-27 kg =1.6605*10^-27*(3*10^8)^2 / 1.6*10^-19 =934.053 MeV$

$(E =m*c^2)$

$Energy = 1881.27 +942.142 -2817.143 = 6.269 MeV$

#### Coil Torque

A planar solenoid with 4 turns and a surface of $4.75*10^{-3} m^2$ has a current of $i=1.58 A$ in the xz plane as shown.  This coil is situated in an external magnetic field $B = (1.86i – 3.79j – 3.36k) mT$

(a) For this system please find the magnetic potential energy.

b) Please find the magnetic torque of magnetic field $B$ on the solenoid. What is the magnetic potential energy of the coil-magnetic field system?

The magnetic energy of the system is just

$U = -m*B$

where m is the magnetic moment vector of the coil, B is the external magnetic field and * is the dot (scalar) product between vectors.

For a single loop of current I the magnetic moment is given by the equation:

(http://en.wikipedia.org/wiki/Magnetic_moment#Current_loop_representation)

$m =I*S$

Therefore for the entire solenoid:

$m =N*I*S =4*1.58*4.75*10^{-3} =30.02*10^{-3} =3*10^{-2} A*m^2$

The direction of $m$ is perpendicular to the surface of the loop:

$m = 0.03*j A*m^2$

Now the magnetic energy is just

$U = -(0.03*j) *(1.86*i -3.79*j -3.36*k)*10^{-3} =0.03*3.79*10^{-3} =0.1137*10^{-3} J$

b)

The magnetic torque is defined as

(http://en.wikipedia.org/wiki/Magnetic_moment#Definition)

$T = m x B$

$T =\begin{vmatrix} i& j& k&\\ 0& 0.03& 0& *10^{-3}\\1.86& -3.79& -3.36\end{vmatrix}=$
$=(-3.36*0.03*i -1.86*0.03*k)*10^{-3} = – (0.1008*i +0.0558*k)*10^{-3}; N*m$