A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density at the center of the loop.

Respuesta :

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

[tex]B=\dfrac{\mu_oI}{2R}[/tex]

[tex]B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}[/tex]

B = 0.000918 T

or

[tex]B=9.18\times 10^{-4}\ T[/tex]

(b) Energy density at the center of the loop is given by :

[tex]U=\dfrac{B^2}{2\mu_o}[/tex]

[tex]U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}[/tex]

[tex]U=0.335\ J/m^3[/tex]

Hence, this is the required solution.