Question

Monday Homework Problem 8.2 In the design of a special electric circuit, a certain component will be destroyed if exposed to a magnetic field greater than 7.5 × 10−6T. Beside this component is a long, thin, straight copper wire. If the wire is 5.0mm from the component, what is the maximim current that this wire may carry before the component is destroyed?

Answer 1

To determine the maximum current that the wire may carry before the component is destroyed, we use Ampere's law to calculate the magnetic field produced by the wire at the distance of 5.0mm from the component. Using the formula, the maximum current is found to be 2.36 A.

Step-by-step explanation:

To determine the maximum current that the wire may carry before the component is destroyed, we need to use Ampere's law which states that the magnetic field produced by a long straight wire is directly proportional to the current and inversely proportional to the distance.

Ampere's law can be written as B = μ0I / 2πr, where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance.

In this case, the wire is 5.0mm (0.005m) from the component. Given that the component will be destroyed if exposed to a magnetic field greater than 7.5 × 10-6T, we can rearrange the equation to solve for I:

I = (2πrB) / μ0

Substituting the values, we have I = (2 × 3.14 × 0.005 × 7.5 × 10-6) / (4π×10-7) = 2.36 A.

Therefore, the maximum current that the wire may carry before the component is destroyed is 2.36 A.