Question

A circular loop of wire 75 mm in radius carries a current of 113 A. Find the (a) magnetic field strength and (b) energy density at the center of the loop.

Answer 1

The magnetic field strength is 9.47 ×10⁻⁴ T

The energy density at the center of the loop is 0.36 J/m³

Calculating Magnetic field strength & Energy density

From the question, we are to find the magnetic field strength

The magnetic field strength of a loop can be calculated by using the formula,

$B = \frac{\mu_{0} I}{2R}$

Where  B is the magnetic field strength

$\mu_{0}$ is the permeability of free space $(\mu_{0}=4\pi \times 10^{-7} \ N/A^{2})$

$I$ is the current

and R is the radius

From the give information,

$R = 75 \ mm= 75 \times 10^{-3} \ m$

and $I = 113 \ A$

Putting the parameters into the formula, we get

$B = \frac{4\pi \times 10^{-7} \times 113}{2 \times 75 \times 10^{-3} }$

$B = 9.47 \times 10^{-4} \ T$

Hence, the magnetic field strength is 9.47 ×10⁻⁴ T

Now, for the energy density

Energy density can be calculated by using the formula,

$u_{B} = \frac{B^{2} }{2\mu_{0} }$

Where $u_{B}$ is the energy density

Then,

$u_{B}= \frac{(9.47\times 10^{-4} )^{2} }{2 \times 4\pi \times 10^{-7} }$

$u_{B} = 0.36 \ J/m^{3}$

Hence, the energy density at the center of the loop is 0.36 J/m³

Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557