Calculate the energy associated with the magnetic field of a 213-turn solenoid in which a current of 1.77 A produces a magnetic flux of 3.83 time sign 10^-4 T.m^2 in each turn.
here,
magnetic fluc , phi = 3.83 * 10^-4 T.m^2
Energy stored in inductor E = 0.5 * L * i^2 .. (i=current)
For long coil L = u0 * A*N^2/l ..
(l=length, A=Area, N=turns)
combining ..
E = 0.5 * (u0 * AN^2/l)i^2 = 0.5 * (u0 * N^2i^2)(A/l) ◄(Eq1)
getting (A/l) ..
B through coil given by B = u0 * Ni/l
Flux through coil B*A = A(u0 * Ni/l) = (A/l)u0 * Ni
From data, flux = {3.83 * 10^-4 *N}
combining ..
(A/l) = {3.83 * 10^-4 }/u0 * Ni = (3.83 * 10^-4) /(u0 N* i)◄(Eq2)
Putting it all together ... comb.Eq1&2
E = 0.5 * (u0 * N * i^2)(3.83 * 10^-4 /u0 * i*N)
E = 0.5 * (N * i)(3.83 * 10^-4)
number of turns , N = 213
current , I = 1.77 A
E = 0.5 * ( 213 * 1.77) * ( 3.83 * 10^-4)
E = 0.0722 J
the energy stored is 72.2 mJ
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