The value of the horizontal component of the | vedclass

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Question

(a)Horizontal component ${B_H} = B\cos \phi $
Total intensity of earth magnetic field $B = \frac{{{B_H}}}{{\cos \phi }}$
$ = \frac{{1.8 	imes {{10}

Vedclass · Accepted Answer

$2.08 	imes {10^{ - 5}}\,Weber/{m^2}$

Vedclass · Answer

(a)Horizontal component ${B_H} = B\cos \phi $
Total intensity of earth magnetic field $B = \frac{{{B_H}}}{{\cos \phi }}$
$ = \frac{{1.8 	imes {{10}^5}}}{{\cos {{30}^o}}} = \frac{{1.8 	imes {{10}^{ - 5}}}}{{\sqrt 3 /2}} = 2.08 	imes {10^{ - 5}}\,Wb/{m^2}$

The value of the horizontal component of the earth's magnetic field and angle of dip are $1.8 \times {10^{ - 5}}\,Weber/{m^2}$ and $30°$ respectively at some place. The total intensity of earth's magnetic field at that place will be

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