Question

A wire carrying a 25.0 A current bends through a right angle. Consider two 2.00 mm segments of wire, each 3.00 cm from the bend (Figure 1).

a) Find the magnitude of the magnetic field these two segments produce at point P , which is midway between them.
b) Find the direction of the magnetic field at point P

Answer 1

To find the magnitude of the magnetic field at point P, we can use the right-hand rule. The magnitude of the magnetic field is 1.33 × 10⁻⁵ T. The direction of the magnetic field at point P is out of the page.

Step-by-step explanation:

To find the magnitude of the magnetic field at point P, we can use the right-hand rule. When a wire carries a current, it creates a magnetic field around it. The magnitude of the magnetic field at the midpoint between two parallel wires is given by the formula:

B = (μ₀ * I) / (2π * r)

Where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷T·m/A), I is the current through the wire, and r is the distance from the wire to point P. Plugging in the values (I = 25.0 A, r = 3.00 cm), we find the magnitude of the magnetic field is 1.33 × 10⁻⁵ T.

To determine the direction of the magnetic field at point P, we can use the right-hand rule again. If we curl the fingers of our right hand in the direction of the current, the thumb will point in the direction of the magnetic field. Since the two wire segments are perpendicular to each other, the magnetic fields they produce will also be perpendicular. Therefore, both wire segments contribute to the magnetic field at point P in the same direction, which is out of the page.

Answer 2

The magnitude of the magnetic field at the point P is
`1.57*10^(-5)T` and the field is pointing into the page.

Step-by-step explanation:

The general form of a similar question to this is:

`\vec{B} = (\mu _(0) )/(4\pi ) * \oint \frac{Id\vec{l} * \hat{r}}{r^(2) }`

where
`\vec{B}` is the vector of the Magnetic Field,
`\mu _(0)` is the Free Space Permeability Constant (equal to
`4\pi * 10^(-7) (N)/(A^2)`),
`I` is the current, and
`r` is the distance from the segment to the point P. (I will get to the
`d\vec{l} * \hat{r}` term in a bit)

This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.

The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from
`\vec{x} * \vec{y}` to
`xy\sin(\theta)`, where
`\theta` is the angle between the 2 vectors, and
`\hat{r}` is the unit vector of
`r` (i.e.
`\hat{r} = \vec{r}/r`) we can simplify
`d\vec{l} * \hat{r}` to just
`dl \sin(\theta)`.

Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.

Finally,
`(\mu_(0))/(4\pi)` simplifies down to
`10^(-7)`.

This gives us our new equation for the Magnetic Field produced by a single segment at a point:

`B = (Il\sin\theta)/(r^(2))*10^(-7)`

Now we need to find values for
`r` and
`\theta`. Luckily, we are dealing with a 45-45-90 triangle with sides of
`1.5 cm`. This means the distance
`r` is
`(1.5\sqrt2)cm`! Similarly, because it is a 45-45-90 triangle, our
`\theta` is
`45\textdegree`!

Now we can start plugging things in:

`B = ((25A)(2*10^(-3)m)\sin(45\textdegree))/((1.5\sqrt2*10^(-2)m)^2)*10^(-7)(N)/(A^2)`

`B = 7.86^(-6) (N)/(A)` or
`T`

This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.

Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:

`B = 1.57*10^(-5) T`; into the page.