Question

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4. The magnetic field inside a superconducting solenoid is 4.70 T. The solenoid has an inner diameter of 6.20 cm and a length of 26.0 cm.

(a) Determine the magnetic energy density in the field.

uB = J / m3

(b) Determine the energy stored in the magnetic field within the solenoid.

UB = kJ

5.–/2 pointsPSE6 32.P.032.

At t = 0, an emf of 540 V is applied to a coil that has an inductance of 0.820 H and a resistance of 30.0 .

(a) Find the energy stored in the magnetic field when the current reaches half its maximum value.

J

(b) After the emf is connected, how long does it take the current to reach this value?

ms

6.–/1 points PSE6 32.P.046.

A 0.95 µF capacitor is charged by a 44.0 V power supply. The fully-charged capacitor is then discharged through a 9.5 mH inductor. Find the maximum current in the resulting oscillations.

Answer 1

4 a) Let us first calculate the volume of the solenood which is given by = π d²/4 × L = 7.84 × 10^-4 m³.

Magnetic energy is given by the formula = B²/2u_o

B = u₀ n I

Hence magnetic energy =uo n² I²/2= 4.24 × 10^-8 J

Hence Magnetic energy density = Magnetic Energy /Volume

= 5.4 × 10^-5 J/m³

b) Energy stored in the magnetic field = 4.24 × 10^-8 J

2) Maximum current value = E/R = 540/30 A = 18 A

a)Energy stored in the magnetic field = 1/2 L I^2 = 33.21 J

b)

When current is half of maximum value

Then e ​​​​​​-Rt/L = 1/2

Hence t = 0.089 sec

c) Maximum current in LC oscillation is given by

wq₀ Where w =√(1/LC) and qo = maximum charge = CE

Max current = 1.04 × 10^-5 ×4.04 × 10^(-5)

4.2 × 10^(-10) A