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.

Question

asked Jul 21, 2020 15.3k views

1 Answer

Varun Maggo · Answer 1 · 2020-07-26T06:50:44+0000

Step-by-step 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 :

$B=(\mu_oI)/(2R)$

$B=(4\pi * 10^(-7)* 114)/(2* 0.078)$

B = 0.000918 T

or

$B=9.18* 10^(-4)\ T$

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

$U=(B^2)/(2\mu_o)$

$U=((0.000918)^2)/(2* 4\pi * 10^(-7))$

$U=0.335\ J/m^3$

Hence, this is the required solution.