On a clear day at a certain location, a 119-v/m vertical electric field exists near the earth's surface. at the same place, the earth's magnetic field has a magnitude of 5.050 10-5 t.
(a) compute the energy density of the electric field. nj/m3
(b) compute the energy density of the magnetic field. µj/m3

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field =

          = 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field =

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

inverse v=i*r

nice nice

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.