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Magnetic Field Due to a Straight Current Carrying Conductor

A straight ve...

Question

A straight vertical conductor carries a current. At a point $5 c m$ due north of it, the magnetic induction is found to be $20 T$ due east. The magnitude of magnetic induction at a point $10 c m$ east of it will be_____

A

20 T

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B

40 T

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C

5 T

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D

10 T

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Solution

The correct option is C $10 T$
$B α \frac{1}{r}$
$\frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}}$
$B_{2} = 20 \times \frac{5}{10}$
$= 10 T$

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