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# As you know, the magnetic north pole of the earth does not

ISBN: 9780321856562 45

## Solution for problem 25P Chapter 11

Introduction to Electrodynamics | 4th Edition

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Problem 25P

Problem 25P

As you know, the magnetic north pole of the earth does not coincide with the geographic north pole—in fact, it’s off by about 11? Relative to the fixed axis of rotation, therefore, the magnetic dipole moment of the earth is changing with time, and the earth must be giving off magnetic dipole radiation.

(a) Find the formula for the total power radiated, in terms of the following parameters: Ψ (the angle between the geographic and magnetic north poles), M (the magnitude of the earth’s magnetic dipole moment), and ω (the angular velocity of rotation of the earth). [Hint: refer to Prob. 11.4 or Prob. 11.11.]

(b) Using the fact that the earth’s magnetic field is about half a gauss at the equator, estimate the magnetic dipole moment M of the earth.

(c) Find the power radiated.

(d) Pulsars are thought to be rotating neutron stars, with a typical radius of 10 km, a rotational period of 10−3 s, and a surface magnetic field of 108 T. What sort of radiated power would you expect from such a star?20

Reference Prob. 11.4

A rotating electric dipole can be thought of as the superposition of two oscillating dipoles, one along the x axis and the other along the y axis (Fig. 11.7), with the latter out of phase by 90˚:

Using the principle of superposition and Eqs. 11.18 and 11.19 (perhaps in the form suggested by Prob. 11.2), find the fields of a rotating dipole. Also find the Poynting vector and the intensity of the radiation. Sketch the intensity profile as a function of the polar angle θ, and calculate the total power radiated. Does the answer seem reasonable? (Note that power, being quadratic in the fields, does not satisfy the superposition principle. In this instance, however, it seems to. How do you account for this?)

Reference Prob. 11.11

A current I (t) flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t), of the loop.

Figure 11.8

Equation 11.60

Step-by-Step Solution:

Step 1 of 5</p>

1)First we have to derive the formula to find the power in terms of

2)We have to find the earth magnetic dipole moment.

3) We have to calculate the power radiating in different situations.

Step 2 of 5</p>

Here, M is the magnetic moment of dipole

angular velocity

The power radiated will be twice that of an oscillating magnetic dipole with dipole moment of amplitude

Therefore

Step 3 of 4

Step 4 of 4

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