A current of 4.83 A in a long straight wire produces a magnetic field of 1.15 $μ$ $mu$ T at a certain distance from the wire. What is this distance?

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Answer 1 · 2017-05-25T09:03:00

$answer d=0.84 m$ $"answer d=0.84 m"$

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$B = K \cdot \frac{2 i}{d}$ $B=K*(2i)/d$

$Firstly,we should convert the μ T to Tesla. 1 μ T = 10^{- 6} T$ $"Firstly,we should convert the "mu T" to Tesla. " 1muT=10^(-6)T$

$K = 10^{- 7}, i=4.83 A , B=1.15 \cdot 10^{- 6} T$ $K=10^(-7)" , i=4.83 A , B=1.15"*10^(-6)T$

$1.15 \cdot 10^{- 6} = 10^{- 7} \frac{(2 \cdot 4.83)}{d}$ $1.15*10^(-6)=10^(-7)((2*4.83))/d$

$d=((2*4.83*cancel(10^(-7))))/(1.15*cancel(10^(-6))$

$d=(2*4.83*10^(-1))/(1.15)$

$d=(9.66*10^(-1))/(1.15)$

$d=(0.966)/(1.15)$

$d=0.84 m$

A current of 4.83 A in a long straight wire produces a magnetic field of 1.15 muμmuT at a certain distance from the wire. What is this distance?