# ​​Q.   A planar coil having12 turns carries 15A current. The coil is oriented with respect to the uniform magetic field  B =0.2 i^ T. such that its directed area is A = -0.04 i^ m2.    The potential energy of the coil in the given orientation is what ?

deepak
84 Points
7 years ago
as you know a coil in a magnetic field only experiences torque, and

$\overrightarrow{T}$ = $\overrightarrow{M}$ X $\overrightarrow{B}$   where T is torque , M is magnetic dipole moment and B is magnectic field

we know that
$\overrightarrow{M}$ = NI$\overrightarrow{A}$  where N is the no of turns i is the current and A is the areal vector
$\overrightarrow{M}$=12x15x(-0.04 $\widehat{i}$)

also $\overrightarrow{B}$=0.2$\widehat{i}$

now consider ,
$\overrightarrow{T}$ = $\overrightarrow{M}$ X $\overrightarrow{B}$    => T=MBsin $\Theta$

we know that workdone by torque,

W.D = $\int T*d\Theta$

W.D = $\int MB sin\Theta * d\Theta$

therefore

W.D=− MB cos$\Theta$
W.D= −$\overrightarrow{M}$.$\overrightarrow{B}$

substituting the value of M and B

W.D = P.E = +12x15x0.2x0.004 = 0.144 J

hope this was usefull XD