Question 35: Magnetism and Matter

Question

The vertical component of the earth's magnetic field is $$6 	imes 10^{-5} \mathrm{~T}$$ at any place where the angle of dip is $$37^{\circ}$$. The earth's resultant magnetic field at that place will be $$\left(ight.$$Given $$\left.	an 37^{\circ}=\frac{3}{4}ight)$$

Accepted Answer

$$	herefore$$ $$\sin 37^\circ  = {{{B_v}} \over {{B_{net}}}}$$
$$ \Rightarrow {B_v} = {B_{net}}\sin 37^\circ $$
$$ \Rightarrow {B_{net}} = {{{B_v}} \over {\sin 37^\circ }}$$
$$ = {{6 	imes {{10}^{ - 5}}} \over {{3 \over 5}}}$$
$$ = 10 	imes {10^{ - 5}}\,T$$
$$ = 1 	imes {10^{ - 4}}\,T$$

Answer

$$8 	imes 10^{-5} \mathrm{~T}$$

Answer

$$6 	imes 10^{-5} \mathrm{~T}$$

Answer

$$5 	imes 10^{-4} \mathrm{~T}$$

Answer

$$1 	imes 10^{-4} \mathrm{~T}$$

Magnetism and Matter

The vertical component of the earth's magnetic field is $6 \times 10^{-5} \mathrm{~T} at any place where the angle of dip is 37^{\circ}. The earth's resultant magnetic field at that place will be \left(\right.Given \left.\tan 37^{\circ}=\frac{3}{4}\right)

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