Calculating Magnetic Field Strength from Wire & Rod Currents

[decamij]

I'm kinda stuck on this question:

1. Calculate the magnitude ofthe magnetic f ield at a point midway between two
long, parallel wires that are 1 .0 m apar t and have currents of1 0.0 A and 20.0 A,
respectively, ifthe currents are (a) in opposite directions and (b) in the same
direction.

The answer is
(a) 1.2 x 10^-5 T
(b) 4.0 x 10^-6 T

2. Also, this question (i thought it was easy, but i kept on getting 4.0 x 10^-5, but the answer is 1.0 x 10^-5).

A long, solid, copper rod has a circular cross-section ofdiameter 10.0 cm. The
rod has a current of5.0 A, uniformly distributed across its cross-section.
Calculate the magnetic field strength 2.5cm from the centre of the rod.


P.S. At this point, the only equation i am using is
B = (mew)oI
2(pi)r
Where (mew)o is the is a proportionality constant,called the permeability of free space.

 

[spacetime]

(1)
Magnetic field due to an infinite wire is given by
\frac{\mu_0i}{2 \pi r}.
And the direction is the same as the direction of cross product of a vector along the current and another in the direction of r.

So, in one case the magnetic fields reinforce each other and in the other, are in the oppsite direction.

(2)
You need to replace the current by \frac{Ir^2}{R^2} since the contribution to the magnetic field is only from the portion of current outside the 2.5 cm cylinder.


spacetime
www.geocities.com/physics_all

 

[wais]

It has a value of 4pi x 10^-7 Tm/A.

First, let's address the calculation for the magnetic field strength between two parallel wires. The key to solving this problem is to use the right-hand rule, which states that if you point your right thumb in the direction of the current in one wire, your fingers will curl in the direction of the magnetic field.

(a) In this case, the currents are in opposite directions, so the magnetic fields created by each wire will also be in opposite directions. Using the right-hand rule, we can see that the magnetic field at the point midway between the wires will be pointing downwards. To calculate the magnitude of the field, we can use the equation B = (mew)oI/2(pi)r, where (mew)o is the permeability of free space, I is the current in the wire, and r is the distance from the wire. Plugging in the values, we get B = (4pi x 10^-7 Tm/A)(10.0 A)/2(pi)(0.5 m) = 1.2 x 10^-5 T.

(b) If the currents are in the same direction, the magnetic fields created by each wire will also be in the same direction. This time, using the right-hand rule, we can see that the magnetic field at the point midway between the wires will be pointing upwards. Plugging in the values into the same equation, we get B = (4pi x 10^-7 Tm/A)(30.0 A)/2(pi)(0.5 m) = 4.0 x 10^-6 T.

For the second question, we can use the same equation, but we need to take into account the fact that the current is distributed across the entire cross-section of the rod. This means we need to use the total current of the rod, which is given as 5.0 A. The distance from the center of the rod to the point where we want to calculate the magnetic field is 2.5 cm, which is equal to 0.025 m. Plugging in the values, we get B = (4pi x 10^-7 Tm/A)(5.0 A)/2(pi)(0.025 m) = 1.0 x 10^-5 T.

It's important to note that in this problem, the diameter of the rod is not relevant in