It is proposed to store 1.00 kW·h = 3.60 × 10⁶ J of electrical energy in a uniform magnetic field with a magnitude of 0.600 T.

What volume (in a vacuum) must the magnetic field occupy to store this amount of energy?

A) V = 6.00 × 10⁶ m³
B) V = 6.00 m³
C) V = 6.00 × 10⁻⁶ m³
D) V = 6.00 × 10¹⁰ m³

The volume required to store 1.00 kW·h of energy in a 0.600 T magnetic field is 6.00 m³, calculated by first finding the magnetic energy density and then dividing the total energy by this density.

Explanation:

The energy density of a magnetic field is given by the formula WB = B2 / (2μ0), where WB is the energy density in joules per cubic meter (J/m3), B is the magnetic field strength in teslas, and μ0 is the permeability of free space (approximately 4π × 10-7 N/A2). To find the volume required to store a given energy, we need to divide the total energy by the energy density.

First, calculate the energy density using the provided magnetic field strength:

WB = (0.600 T)2 / (2 × 4π × 10-7 N/A2)

Then, divide the total energy by this energy density to get the volume:

V = 3.60 × 106 J / WB

Substitute the values and solve for V, which results in the correct answer of V = 6.00 m3.