Question

The magnetic field inside a superconducting solenoid is 4.50 t. the solenoid has an inner diameter of 6.20 cm and a length of 26.0 cm. determine (a) the magnetic energy density in the field and (b) the energy stored in the magnetic field within the solenoid.

Answer 1

Answer:

(a) $u_B=8057218.9\ J/m^3$

(b) $U=6324.58\ J$

Explanation:

(a) The magnetic energy density in the field is given by:

$u_B=\frac{B^2}{2\mu_o}$   ( 1 )

B: magnitude of the magnetic field

μo: magnetic permeability of vacuum = 4pi*10^-7 T/A

you replace the values of B and μo in the equation (1):

$u_B=\frac{(4.50T)^2}{2(4\pi *10^{-7}T/A)}\\\\u_B=8057218.9\ J/m^3$

(b) The energy stored in the magnetic field of the solenoid:

$U=\frac{B^2}{2\mu_o}V$  ( 2 )

V: volume of the solenoid

$V=\pi (\frac{0.062m}{2})^2(0.26m)=7.84*10^{-4}m^3$

Next, you replace V in the equation (2). Furthermore, you can use the result obtained in (a) for u_B:

$U=u_BV=(8057218.9J/m^3)(7.84*10^{-4}m^3)\\\\U=6324.58\ J$

Answer 2

The magnetic energy density is 8.057 × 10⁶ J/m³. The energy stored in the magnetic field within the solenoid is 6.324 × 10³ J

The magnetic energy density of the superconducting solenoid can be expressed by using the relation:

$\mathbf{\mu_{_B} = \dfrac{1}{2} \dfrac{B^2}{\mu_o} }$

where;

• B = magnetic field
• $\mathbf{\mu_o}$ = magnetic permeability

$\mathbf{\mu_{_B} = \dfrac{1}{2} \Big( \dfrac{(4.50)^2}{4 \pi \times 10^{-7} }\Big)}$

$\mathbf{\mu_{_B} =8057218.994 \ J/m^3}$

$\mathbf{\mu_{_B} =8.057 \times 10^6 \ J/m^3}$

The energy stored in the magnetic field within the solenoid is determined by using the formula:

$\mathbf{U = \mu_B \times A \times l}$

where;

• $\mathbf{\mu_B}$ = magnetic energy density
• A = area = πr²
• L = length

∴

$\mathbf{U = 8.057 \times 10^6 \times \pi(\dfrac{6.2 \times 10^{-2} \ m}{2} )^2 \times (26\times 10^{-2} \ m)}$

$\mathbf{U = 6324.4 \ J}$

U = 6.324 × 10³ J

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