The magnetic field due to a short magnet at a point on its...

The magnetic field due to a short magnet at a point on its axis at a distance $\mathrm{x} \mathrm{cm}$ from the middle point of the magnet is 200 gauss. The magnetic field at a point on the neutral axis at a distance $\mathrm{x} \mathrm{cm}$ from the middle of the magnet is gauss. (a) 100 (b) 400 (c) 50 (d) 200

Key Concepts

Magnetic Dipole

A magnetic dipole is a simplified model used to describe the magnetic field of a magnet, treating it as if it were composed of two equal and opposite magnetic charges separated by a small distance. This approximation is particularly useful when analyzing the field at distances that are large compared to the size of the magnet.

Axial and Equatorial Magnetic Fields

The magnetic field produced by a dipole has different values along different axes. Specifically, the field along the dipole’s axis (axial field) is stronger than the field along the equatorial or neutral plane (which is perpendicular to the axis). In the ideal dipole model, the axial field is typically twice the magnitude of the equatorial field at the same distance from the center of the dipole.

Transcript

0:00 Hi there.

00:01 So for this problem, the magnetic field is 40 centimeters away from a long string wire carrying a current that is equal to the current is equal to two umpers.

00:18 And the magnetic field, the magnitude of the magnetic field is equal to one micro tesla.

00:27 And the distance, well, the distance of separation is going to be 40 centimeters.

00:40 Let me just put that in here.

00:44 Okay, so we're told for par a.

00:50 At what distance is it 0 .1 micro -tesla.

00:57 So we need to find when the magnitude of the magnetic field is equal to 0 .100 microtesla.

01:12 Okay, so from the equation of the magnetic field that we know that is mu -sub -0, which is a constant times the current divided by 2 times pi times the radius or the separation, we observed that the field is inversely proportional to the distance from the conductor.

01:34 So the field will have one tenth its original value of the distance is increased by a factor of 10.

01:43 Then the required distance that we are going to call the distance prime is 10 times the distance are that we are given.

01:53 And the distance art that we are given is 40 centimeters that in meters is 0 .4 meters.

02:01 So from this, we obtain that we need to put ourselves into a distance of four meters in order to measure that 0 .1 micro -tesla.

02:12 So that's a solution for part a of this problem.

02:18 Now, for part b, we are asked about at one instant the two conductors in a loan household exemption court current and equal to ampers currents in the opposite direction.

02:31 Now the wires are 3 millimeters apart, 3 millimeters apart.

02:40 And we need to find the magnetic field 40 centimeters away from the middle of the straight corn in the plane of the two wires.

02:54 So with that set, so we know that at a point in the plane of the conductors and 40 centimeters from the center of the core is located located 39 .85 centimeters from the nearer and 40 .15 centimeters from the far wire.

03:28 Since the currents are in opposite directions, we are there so the contributions to the net field is going to be that the net field in this case is going to the sum of these two wires, one minus the order.

03:48 So we're going to have that the net magnetic field is going to be miso zero times the current.

03:56 The current is the same for both over two times pi times one over the radius one minus one over the radius two.

04:04 And we substitute the values in here.

04:07 So we will have that this is.

04:08 Musup 0 is 4 times pi times 10 to the minus 7, tesla times meters per ampere.

04:15 And this times the current, which is 2 ampers, and this divided by 2 times pi.

04:21 And all of this divided by 1 over the radius 1, which is 0 .300 and 98, 5 meters.

04:36 We pass this from centimeters to meters and minus one, the radius 2, that is going to be 0 .40 -15 meters.

04:46 So from this, holding from this, we will find that the net magnetic field is equal to 7 .5 times 10 to the minus 9 tesla, that we can also write as 7 .5 nanotesla.

05:04 So that's a solution for part b of this problem...

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