Question

The magnetic field 40.0 cm away from a long, straight wirecarrying current 2.00 a is 1.00 mt.

At what distance isit 0.100 mt? what if? at one instant, the two conductorsin a long household extension cord carry equal 2.00-acurrents in opposite directions. the two wires are 3.00 mmapart. find the magnetic field 40.0 cm away from the middleof the straight cord, in the plane of the two wires.

Answer 1

The new distance at which the magnetic field is 0.100 mT is calculated to be 4.0 meters away from a long, straight current-carrying wire. The net magnetic field is zero at a point 40.0 cm from the midpoint between two conductors carrying equal but opposite currents and separated by 3.00 mm.

Step-by-step explanation:

The magnetic field strength exerted by a long, straight current-carrying wire at a certain distance can be calculated using Ampere's Law. The magnetic field B created by a long straight wire carrying a constant current I has a magnitude given by B = (μ_0 * I) / (2π * r), where μ_0 is the permeability of free space (4π x 10-7 T·m/A), I is the current, and r is the distance from the wire.

Using this relationship, if the magnetic field is 1.00 mT at 40.0 cm (0.4 m) from the wire and decreases to 0.100 mT, to find the new distance d at which the magnetic field is 0.100 mT, you would set up a proportion because the magnetic field strength is inversely proportional to the distance from the wire. So, if B1/B2 = d2/d1, then d = (B1/B2) * d1 = (1.00 mT / 0.100 mT) * 0.4 m = 4.0 m away from the wire.

For the situation with two conductors carrying equal but opposite currents of 2.00 A and separated by 3.00 mm, the magnetic fields due to each wire at a point 40.0 cm from the midpoint between them will be opposite in direction and thus will cancel each other out, resulting in a net magnetic field of zero at that point.