If the angles of dip two places are 30-circ- and 45-circ- respectively- then the ratio of horizontal components of earth-s magnetic field the two places will besqrt -3- -sqrt -2-1 - sqrt -2-1 - 21 - sqrt -3-

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Toppr Admin · Accepted Answer

The horizontal component of earth-apos-s magnetic field is given by - H-Rcos-x3B4-where- -x3B4- is the angle of dip so H1-Rcos30oand H2-Rcos45o-x2234-H1H2-Rcos30oRcos45o-x221A-3-21-x221A-2-x221A-3-x221A-2

If the angles of dip at two places are $30^{\circ}$ and $45^{\circ}$ respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be
$\sqrt{3} : \sqrt{2}$
$1 : \sqrt{2}$
$1 : 2$
$1 : \sqrt{3}$

The horizontal component of earth's magnetic field is given by , $H = R cos δ$
where, $δ$ is the angle of dip so $H_{1} = R cos 30^{o}$
and $H_{2} = R cos 45^{o}$
$∴ \frac{H_{1}}{H_{2}} = \frac{R cos 30^{o}}{R cos 45^{o}} = \frac{\sqrt{3} / 2}{1 / \sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$

If the angles of dip at two places are 30∘ and 45∘ respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be√3:√21:√21:21:√3

The horizontal component of earth's magnetic field is given by , H=Rcosδwhere, δ is the angle of dip so H1=Rcos30oand H2=Rcos45o∴H1H2=Rcos30oRcos45o=√3/21/√2=√3√2

If the angles of dip at two places are $30^{\circ}$ and $45^{\circ}$ respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be
$\sqrt{3} : \sqrt{2}$
$1 : \sqrt{2}$
$1 : 2$
$1 : \sqrt{3}$

The horizontal component of earth's magnetic field is given by , $H = R cos δ$
where, $δ$ is the angle of dip so $H_{1} = R cos 30^{o}$
and $H_{2} = R cos 45^{o}$
$∴ \frac{H_{1}}{H_{2}} = \frac{R cos 30^{o}}{R cos 45^{o}} = \frac{\sqrt{3} / 2}{1 / \sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$