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What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in air? Give your answer in units of tesla.

User Anuni
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1 Answer

4 votes

Answer:

The magnetic flux density is
2.11*10^(-6)\ T

Step-by-step explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field


B =(\mu_(0)I)/(2r)

Where,

r = radius

I = current

Put the value into the formula


B =(4\pi*10^(-7)*3.8)/(2*\pi*0.36)


B=2.11*10^(-6)\ T

Hence, The magnetic flux density is
2.11*10^(-6)\ T

User Textual
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