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Chapter 20: Problem 77

How much energy due to Earth's magnetic field is present in $1.0 \mathrm{m}^{3}$ of space near Earth's surface, at a place where the field has magnitude \(0.45 \mathrm{mT} ?\)

Short Answer

Expert verified
Answer: The energy is approximately 1.91 脳 10鈦烩伔 J.

Step by step solution

01

Calculate the energy density (u) in the magnetic field

Using the formula for the energy density in a magnetic field, we can plug in the given value for the magnetic field magnitude and the permeability of free space. \(B = 0.45 \times 10^{-3} \mathrm{T}\) (converting mT to T) \(u = \frac{1}{2(4\pi\times10^{-7})}(0.45\times10^{-3})^2\)
02

Calculate the energy density (u) value

Solving for u: \(u \approx 1.91\times10^{-7}\frac{\mathrm{J}}{{m}^{3}}\) Now, we have the energy density in the magnetic field.
03

Calculate the energy present in the volume of the space

We can multiply the energy density (u) by the given volume to find the total energy present in this volume of space. \(E = u \cdot V \) \(E = (1.91\times10^{-7}\mathrm{\frac{J}}{{m}^{3}}) \cdot (1.0\mathrm{m}^{3})\)
04

Calculate the energy (E) value

Solving for E: \(E \approx 1.91\times10^{-7}\mathrm{J}\) Therefore, the energy due to Earth's magnetic field present in \(1.0\mathrm{m}^{3}\) of space near Earth's surface, at a place where the field has a magnitude \(0.45\mathrm{mT}\), is approximately \(1.91\times10^{-7}\mathrm{J}\).

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