What is the flux density at a point 3 cm from the current carrying conductor, when there is a current of 25 A in a wire: (u0= 4𝜋 x10-7)
Correct answer: B. 1.67 x 10^-4T
- A. 0.23 x 10^-1T
- B. 1.67 x 10^-4T
- C. 2.99 x 10^-6T
- D. 3.63 x 10^-8T
- E. 999 x 10^-7T
Explanation
The correct answer is 1.67x10-4T. To find the magnetic flux density (B) at a distance (r) from a long straight wire carrying current (I), you can use the formula:B = (μ0 * I) / (2 * π * r).Substituting in the values: μ0 = 4π x 10-7 T·m/A, I = 25 A, and r = 0.03 m (3 cm), we get:B = (4π x 10-7 * 25) / (2 * π * 0.03) = 1.67 x 10-4T.Other options are incorrect because:0.23x10-1T is too high, not matching the expected results.2.99x10-6T is too low compared to the calculations.3.63x10-8T is orders of magnitude smaller than expected.999x10-7T converts to an unreasonably high value of 0.0999 T.
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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