Certainly! Let's begin by isolating [tex]\( J \)[/tex] in the equation [tex]\( T = \frac{1}{2AJ^2} + \text{trs} \)[/tex].
1. Start with the given equation:
[tex]\[ T = \frac{1}{2A J^2} + \text{trs} \][/tex]
2. Subtract [tex]\(\text{trs}\)[/tex] from both sides to isolate the term involving [tex]\( J \)[/tex]:
[tex]\[ T - \text{trs} = \frac{1}{2A J^2} \][/tex]
3. Now, we want to get rid of the fraction. Multiply both sides by [tex]\( 2A J^2 \)[/tex] to clear the denominator:
[tex]\[ (T - \text{trs})(2A J^2) = 1 \][/tex]
4. Divide both sides by [tex]\( 2A (T - \text{trs}) \)[/tex] to isolate [tex]\( J^2 \)[/tex]:
[tex]\[ J^2 = \frac{1}{2A (T - \text{trs})} \][/tex]
5. Finally, take the square root of both sides to solve for [tex]\( J \)[/tex]:
[tex]\[ J = \pm\sqrt{\frac{1}{2A (T - \text{trs})}} \][/tex]
As a result, we have two solutions for [tex]\( J \)[/tex]:
[tex]\[ J = -\sqrt{\frac{1}{2A (T - \text{trs})}} \][/tex]
[tex]\[ J = \sqrt{\frac{1}{2A (T - \text{trs})}} \][/tex]
