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. Enhance your understanding with multiple choice questions, each equipped with hints and explanations. Prepare effectively for high performance on your exam!

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.

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy