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Standard XII

Physics

Earth's Magnetism

A magnetic co...

Question

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45^{o}$ , and $40$ times per minute where the dip is $30^{o}$ . If $B_{1}$ and $B_{2}$ are respectively the total magnetic field due to the earth at the two places, then the ratio $B_{1} / B_{2}$ is best given by :

2.2

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1.8

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0.7

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3.6

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Solution

The correct option is C $0.7$
$f_{1} = \frac{1}{2 π} \sqrt{\frac{μ B_{1} cos 45^{o}}{I}}$
$f_{2} = \frac{1}{2 π} \sqrt{\frac{μ B_{2} cos 30^{o}}{I}}$
$\frac{f_{1}}{f_{2}} = \sqrt{\frac{B_{1} cos 45^{o}}{B_{2} cos 30^{o}}}$
$∴ \frac{B_{1}}{B_{2}} = 0.7$

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A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45o, and 40 times per minute where the dip is 30o. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is best given by :

The correct option is C 0.7f1=12π√μB1cos45oIf2=12π√μB2cos30oIf1f2=√B1cos45oB2cos30o∴B1B2=0.7

The correct option is C $0.7$
$f_{1} = \frac{1}{2 π} \sqrt{\frac{μ B_{1} cos 45^{o}}{I}}$
$f_{2} = \frac{1}{2 π} \sqrt{\frac{μ B_{2} cos 30^{o}}{I}}$
$\frac{f_{1}}{f_{2}} = \sqrt{\frac{B_{1} cos 45^{o}}{B_{2} cos 30^{o}}}$
$∴ \frac{B_{1}}{B_{2}} = 0.7$