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Angle of dip delta and latitude lambda on earth's surface are related as

Answer»

`tan delta = 2 tan lambda`
`tan delta = cot lambda`
`tan delta = (tan lambda)/(2)`
`tan delta = tan lambda`

Solution :`B_(R) = (mu_(0))/(4pi) (2M cos theta)/(r^(3))`
`B_(theta) =(mu_(0))/(4pi) (M SIN theta)/(r^(3))`
SINCE `theta = 90^(@) + lambda` , thus
`B_(r) = (mu_(0))/(4pi) (2M cos (90 + lambda))/(r^(3)) = - (mu_(0))/(4pi)(2M sin lambda)/(r^(3))`
`B_(theta) = (mu_(0))/(4pi) (M sin (90 + lambda))/(r^(3)) = (mu_(0))/(4pi) (M cos lambda)/(r^(3))`
`therefore tan delta = (B_(V))/(B_H) = 2 tan lambda`


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