Which expression correctly gives the energy density in a linear magnetic material in terms of H?

Study for the NEIEP Magnetism and Electromagnetism (355) exam. Prepare with our interactive quizzes, multiple-choice questions, and detailed explanations. Ace your test and enhance your knowledge on magnetism principles.

Multiple Choice

Which expression correctly gives the energy density in a linear magnetic material in terms of H?

Explanation:
The energy stored per unit volume in a magnetic field is found from the work done to build up the field, which is expressed as u = ∫ H · dB. For a linear, isotropic material, B and H are proportional: B = μ H with μ constant. This gives dB = μ dH, and the energy density becomes u = ∫ from 0 to H of μ H' dH' = (1/2) μ H^2. So the expression that matches this result is u = 1/2 μ H^2. This form also makes physical sense: the energy is zero when the field is zero and increases with H, with μ controlling how much energy is stored. The other forms would yield μ H^2 (or B^2/μ) in a linear material, missing the 1/2 factor that arises from the integral of H with respect to B during the magnetization process. In vacuum, μ would be μ0, and in a linear material it would be μ = μ0 μ_r, but the half-factor remains.

The energy stored per unit volume in a magnetic field is found from the work done to build up the field, which is expressed as u = ∫ H · dB. For a linear, isotropic material, B and H are proportional: B = μ H with μ constant. This gives dB = μ dH, and the energy density becomes u = ∫ from 0 to H of μ H' dH' = (1/2) μ H^2. So the expression that matches this result is u = 1/2 μ H^2.

This form also makes physical sense: the energy is zero when the field is zero and increases with H, with μ controlling how much energy is stored. The other forms would yield μ H^2 (or B^2/μ) in a linear material, missing the 1/2 factor that arises from the integral of H with respect to B during the magnetization process. In vacuum, μ would be μ0, and in a linear material it would be μ = μ0 μ_r, but the half-factor remains.

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