$\begin{array}{1 1} -4.8 \times 10^{-2}\;J \\ 4.8 \times 10^{-2}\;J \\ 48 \times 10^{-2}\;J \\ -48 \times 10^{-2}\;J \end{array} $

Solution :

Given :

$m= 0.32 \;J/T$

$B= 0.15 \;T$

$\theta= 180^{\circ}$

For unstable equilibrium . the angle between the magnetic moment and the magnetic field is $180^{\circ}$

the potential energy of the magnet.

$U= -mB \cos \theta^{\circ}$

$\qquad= 0.32 \times 0.15 \times (-1)$

$(\cos 180^{\circ} =-1) $

$\qquad= =4.8 \times 10^{-2}\;J$

Answer : $ 4.8 \times 10^{-2}\;J $

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