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Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is the magnitude of the magnetic field at a point that is 5.0 cm from each of the wires

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3 votes

Answer:

24μT

Step-by-step explanation:

According to the question the horizontzal component of the field at P are equal and opposite therefore, it will get cancelled.

And vertical components will be added

So, net magnetic field will be
B_(net)


B_(net)= 2Bsinθ


B=(\mu _(0)I)/(2\pi*5*10^(-2))=(2*10^(-7)*5)/(5*10^(-2))


B=2*10^(-5)T

therefore,
B_(net)=
2*2*10^(-5)* [\tex][tex](3)/(5) ( sinθ=3/5)

on calculating we get
B_(net)= 24μT

User JPNagarajan
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