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consider two long, straight, horizontal wires that are oriented so that they are perpendicular to each other. the two wires are displaced vertically from each by 5.8 cm, so that, and at their closest, they are 5.8 cm apart. the top wire carries a current of 78 a, while the bottom one carries 94 a. how strong is the magnetic field (in mt) at a point midway between the two wires, on the vertical line that connects the two points of closest approach?

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Final answer:

To find the magnetic field at a point midway between two perpendicular wires, calculate the individual magnetic fields due to each wire using B = (µ0I)/(2πr), and then determine the resultant by vector addition, considering the orientation and magnitude of the fields from each wire.

Step-by-step explanation:

To solve the mathematical problem of finding the magnetic field at a point midway between two perpendicular wires carrying different currents, we need to use the principle that the magnetic field (B) due to a long straight wire at a distance (r) is given by B = (µ0I)/(2πr), where I is the current and µ0 is the permeability of free space, which has a value of 4π × 10-7 Tm/A. For each wire, we calculate the magnetic field at the point of interest separately and then vectorially add the two fields to get the total field at that point.

The top wire with 78 A will produce a magnetic field pointing into the screen (using the right-hand rule), and the bottom wire with 94 A will produce a magnetic field pointing to the right. Midway between the wires, the magnetic field contributions from each wire will be perpendicular to each other hence the resultant field is found using Pythagoras theorem as B = √(B12 + B22), where B1 and B2 are the magnetic fields due to the top and bottom wires respectively.

After calculating B1 and B2, if the point of interest is midway and at a distance of 2.9 cm from each wire (half of 5.8 cm), the total magnetic field can be found. Once this calculation is done, the magnetic field will be expressed in milliteslas (mT), as required.

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