Question #07f44

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Answer 1 · 2016-04-30T18:19:14

$The net magnetic field at E: - \to B$ $"The net magnetic field at E:"- vec B$

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$for green wire :$ $"for green wire :"$

${\to B}_{X} = \frac{2 \cdot K \cdot I_{1}}{\frac{d}{2}} = \frac{4 \cdot K \cdot I_{1}}{d}$ $vec B_X=(2*K*I_1)/(d/2)=(4*K*I_1)/d$

${\to B}_{X} = \frac{8 K}{d}$ $vec B_X=(8K)/d$

$for red wire :$ $"for red wire :"$

${\to B}_{O} = \frac{2 \cdot K \cdot I_{2}}{\frac{d}{2}} = \frac{2 \cdot K \cdot 1}{\frac{d}{2}}$ $vec B_O=(2*K*I_2)/(d/2)=(2*K*1)/(d/2)$

${\to B}_{O} = \frac{4 K}{d}$ $vec B_O=(4K)/d$

${\to B}_{X} > {\to B}_{O}$ $vec B_X > vec B_O$

$B = \frac{8 K}{d} - \frac{4 K}{d} =$ $B=(8K)/d-(4K)/d=$

$B = \frac{4 K}{d}$ $B=(4K)/d$

$current I_{1} = 2A switched off :$ $"current " I_1="2A switched off :"$

$The net magnetic field is : - \to B$ $"The net magnetic field is :"- vec B$