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It is proposed to store1kW.h=3.6×106Jof electrical energy in a uniform magnetic field with magnitude0.6T. (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube40cmon a side, what magnetic field is required?

Short Answer

Expert verified

a) V=25.1m3

b)B=11.9T

Step by step solution

01

The formulas involved in the given problem

The energy density can be calculated by u=UVor u=B22μ0.

Equate both the equations and find the volume.

02

Calculate the volume

Given thatU=3.6×106J,l=40cmandB=0.6T. Using both the energy density equations

UV=B22μ0V=2μ0UB2V=2×4π×10-7×3.6×1060.62V=25.1m3

The side of the cube isl=40cm. The volume will be 0.064m3.

B=2μ0UVB=2×4π×10-7×3.6×1060.064B=11.9T

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