Chapter 21, Problem 45P

To determine

To find the energy stored in the magnetic field.

Answer to Problem 45P

Solution:

$U=32.4\text{ J}$

Explanation of Solution

Given:

Length of the solenoid, $l=36\text{ cm = 0}\text{.360 m}$

Diameter of the solenoid, $d=2\text{ cm = 0}\text{.02 m}$ d = 2.0 cm = 0.02 m

Radius of the solenoid, $r=1\text{ cm = 0}\text{.01 m}$

Magnetic field, $B=0.72\text{ T}$

Formula used:

The energy stored in the magnetic field

$U=\frac{1}{2}\frac{B}{{\mu }_0}\left(\pi r^{2}l\right)$

Calculation:

Substituting the given values the above relation, we get the energy stored in the magnetic field

$U=\frac{1}{2}\frac{B}{{\mu }_0}\left(\pi r^{2}l\right)=\frac{1}{2}\times \frac{\left(0.72\right)}{4\times \pi \times {10}^{-7}}\left(\pi \times {0.01}^2\times 0.360\right)=32.4\text{ J}$

Therefore, 32.4 J of energy is stored in the magnetic field.