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Magnetic Field Due to a Straight Long Wire

The magnetic ...

Question

The magnetic induction at a point P which is distant 4 cm from a long current carrying wire is $10^{- 3}$ Telsa.The field of induction at a distance 12 cm from the same current would be

$3.33 \times 10^{- 9}$ Tesla

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$1.11 \times 10^{- 4}$ Tesla

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$3 \times 10^{- 3}$ Tesla

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$9 \times 10^{- 2}$ Tesla

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Solution

The correct option is A
$3.33 \times 10^{- 9}$ Tesla

$B = \frac{μ_{0}}{4 π} . \frac{2 i}{r} \Rightarrow \frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{10^{- 3}}{B_{2}} = \frac{12}{4} \Rightarrow B_{2} = 3.33 \times 10^{- 9}$ Tesla

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The magnetic induction at a point P which is distant 4 cm from a long current carrying wire is 10−3 Telsa.The field of induction at a distance 12 cm from the same current would be

The correct option is A 3.33×10−9 Tesla B=μ04π.2ir⇒B1B2=r2r1⇒10−3B2=124⇒B2=3.33×10−9 Tesla

The magnetic induction at a point P which is distant 4 cm from a long current carrying wire is $10^{- 3}$ Telsa.The field of induction at a distance 12 cm from the same current would be

The correct option is A
$3.33 \times 10^{- 9}$ Tesla

$B = \frac{μ_{0}}{4 π} . \frac{2 i}{r} \Rightarrow \frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{10^{- 3}}{B_{2}} = \frac{12}{4} \Rightarrow B_{2} = 3.33 \times 10^{- 9}$ Tesla