A long straight vertical conductor carries a current of 8A in the upward direction. If the horizontal component of earth's magnetic induction is 4×10−5 T . Then at a distance of 4cm to the south of it the resultant magnetic induction in the horizontal plane is,

B=μ0i2πr =2×10−7×84×10−2 =4×10−5T BH=4×10−5T Bnet=B2+BH2 =(4×10−5)2+(4×10−5)2 =32×10−5T Bnet=42×10−5T

A long straight vertical conductor carries a current of 8A in the upward direction. If the horizontal component of earth's magnetic induction is 4×10−5 T . Then at a distance of 4cm to the south of it the resultant magnetic induction in the horizontal plane is,

B=μ0i2πr =2×10−7×84×10−2 =4×10−5T BH=4×10−5T Bnet=B2+BH2 =(4×10−5)2+(4×10−5)2 =32×10−5T Bnet=42×10−5T

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A long straight vertical conductor carries a current of 8A in the upward direction. If the horizontal component of earth's magnetic induction is $4 \times 10^{- 5} T$ . Then at a distance of 4cm to the south of it the resultant magnetic induction in the horizontal plane is,

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2 \times 10^{- 5} T

8 \times 10^{- 5} T

4 \sqrt{2} \times 10^{- 5} T

4 \times 10^{- 5} T

answer is C.

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Detailed Solution

B = \frac{μ_{0} i}{2 π r}

$= \frac{2 \times 10^{- 7} \times 8}{4 \times 10^{- 2}}$

$= 4 \times 10^{- 5} T$

$B_{H} = 4 \times 10^{- 5} T$

$B_{net} = \sqrt{B^{2} + B_{H}^{2}}$

$= \sqrt{{(4 \times 10^{- 5})}^{2} + {(4 \times 10^{- 5})}^{2}}$

$= \sqrt{32} \times 10^{- 5} T$

$B_{net} = 4 \sqrt{2} \times 10^{- 5} T$

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A long straight vertical conductor carries a current of 8A in the upward direction. If the horizontal component of earth's magnetic induction is 4×10−5 T . Then at a distance of 4cm to the south of it the resultant magnetic induction in the horizontal plane is,

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A long straight vertical conductor carries a current of 8A in the upward direction. If the horizontal component of earth's magnetic induction is $4 \times 10^{- 5} T$ . Then at a distance of 4cm to the south of it the resultant magnetic induction in the horizontal plane is,