Ncert Exercises & Solutions

Assume the dipole model for the earth's magnetic field B which is given by,

$B_{v}$ = vertical component of the magnetic field $= \frac{μ_{0}}{4 π} \frac{2 m \cos θ}{r^{3}}$
$B_{H}$ = horizontal component of the magnetic field $= \frac{μ_{0}}{4 π} \frac{msin θ}{r^{3}}$
$θ = 90 °$ = lattitude as measured from the magnetic equator.

Find loci of points for which
(a) |B| is minimum
(b) dip angle is zero and
(c) dip angle is 45°

Hint: Use the formula of the magnetic field due to bar magnet.

(a) Step 1:

B_{V} = \frac{μ_{0}}{4 π} \frac{2 m \cos θ}{r^{3}} . . . (i)

B_{H} = \frac{μ_{0}}{4 π} \frac{msinθ}{r^{3}} . . . (ii)

Squaring both the equations and adding, we get;

\begin{array}{l} B_{V}^{2} + B_{H}^{2} = {(\frac{μ_{0}}{4 π})}^{2} \end{array} \frac{m^{2}}{r^{6}} (4 \cos^{2} θ + \sin^{2} θ)

...iii

B = \begin{array}{l} \sqrt{B_{V}^{2} + B_{H}^{2}} = (\frac{μ_{0}}{4 π}) \end{array} \frac{m}{r^{3}} {(3 \cos^{2} θ + 1)}^{\frac{1}{2}}

From Eq. (i), the value of B is minimum, if

\cos θ = \frac{π}{2}

Thus, the magnetic equator is the locus.

(b) Step 2: Angle of dip,

\begin{array}{l} \tan δ = \frac{B_{V}}{B_{H}} = \frac{\frac{μ_{0}}{4 π} . \frac{2 mcos θ}{r^{3}}}{\frac{μ_{0}}{4 π} \cdot \frac{msin θ}{r^{3}}} = 2 \cot θ . . . (iv) \\ \tan δ = 2 \cot θ \end{array}

For dip angle is zero i.e.,

δ = 0

°

\cot θ = 0

θ = \frac{π}{2}

It means that the locus is again the magnetic equator.

\tan δ = \frac{B_{V}}{B_{H}}

Angle of dip i.e., δ = \pm 45 °

\frac{B_{V}}{B_{H}} = \tan (\pm 45^{\circ})

\frac{B_{V}}{B_{H}} = 1

2 \cot θ = 1 [From Eq . (iv)]

\cot θ = \frac{1}{2}

\tan θ = 2

\Rightarrow θ = \tan^{- 1} (2)

Thus,

θ = \tan^{- 1} (2)

is the locus.

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