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(a) Since distance from wire is much less than the length of the wire, we can assume infinitely long wire equation applies. This gives the magnitude of the magnetic field to be

(b) Writing the field of the current in the wire in \(\text{G}\) we have

Therefore, the field by current is

That is the field of the current will be 4 times as strong.

Using the magnitude formula for field of current in an infinitely long wire,

we get

For comparision, magnetic field of Earth is \(\approx 5.0\times 10^{-5}\text{ T}\text{.}\)

(a) Since other parts of the circuit are far away, we can ignore them and just keep the field of long wires that are nearby point P. We can also make the assumption that the formula for infintely long wire applies. Now, when we apply right hand rule, we find that field of the current in the left wire is into-the-page while the field for the current in the right wire is also into-the-page. Therefore, the two fields will add at point \(\text{P}_1\text{.}\)

The direction is going-into-the-page.

(b) Here, the direction from the nearer wire (right) is out-of-the-page and that from the farther wire (left) is into-the-page. They will oppose each other. Let's call out-of-the-page as positive and then into-the-page will be negative. Continuing to use subscript 1 for the left wire and 2 for the right wire, we get

Since, we found the net as positive, the out-of-page was greater. Therefore, the direction is out-of-the-page.

Since the wire segment is straight, we use a Cartesian coordinate system. We will place the wire symmetrically along the \(z\) axis so that the field point will be in the \(xy\) plane. We will also choose \(x\) axis direction to go through the field point P as shown in Figure 39.1.8.

We choose an infinitesimal element between \(z\) and \(z+dz\text{.}\) This geometry gives us the following for various quantities in the infintesimal form of Biot-Savar law.

Therefore,

Now, we have all the pieces to write the contribution of the steady current in the infinitesimal element at \(z\text{.}\)

We need to integrate this from \(z=-L/2\) to \(z=L/2\text{.}\) Since integrand is an even function in \(z\) and the range of integration is symmetric around zero, we will get double the value from \(z=0\) to \(z=L/2\text{.}\)

Picking the field point on the \(x\) axis gave us direction of the magnetic field towards positive \(y\) axis. You can show that if you pick the field point on positive \(y\) axis, you would get the direction negative \(x\) axis. The direction at arbitrary point will be in the unit vector \(\hat u_\phi\) of the cylindrical coordinate system, i.e., magnetic field in the \(xy\) plane will be circulating in circular path.

Since the wire segment is straight, we use a Cartesian coordinate system. We will place the wire symmetrically along the \(z\) axis. We will use Cartesian coordinates \((x,y,z) \) for location of field point P as shown in Figure 39.1.10. For the segment on the wire we will use symbol \(z'\) in place of letter \(z\text{,}\) which we are using for the field point P.

We choose an infinitesimal element between \(z'\) and \(z'+dz'\text{.}\) This geometry gives us the following for various quantities in the infintesimal form of Biot-Savar law.

Therefore,

Now, we have all the pieces to write the contribution of the steady current in the infinitesimal element at \(z\text{.}\)

Integrating this from \(z'=0\) to \(z'=L\) gives us the following result

This answer can be written more compactly in terms of unit vector \(\hat u_\phi = \left( -\dfrac{y}{s}\, \hat u_x + \dfrac{x}{s}\, \hat u_y \right)\) and angles the two ends of the wire subtends at the field point P as you can see from the figure to the right.

\begin{align*} \sin\,\theta_1 \amp = \dfrac{ L - z}{ \sqrt{ s^2 + (L-z)^2 }} \\ \sin\,\theta_2 \amp = \dfrac{ z}{ \sqrt{ s^2 + (L-z)^2 }} \end{align*}

Therefore, magnetic field at P is compactly written as

where \(s\) is the distance of the field point from the line of the wire. To check the direction we use the right-hand rule, which also gives the direction to be circulating as \(\hat u_\phi\) does.

We will use answer in Checkpoint 39.1.7 and take \(L/s \rightarrow \infty\) to produce the result expected for infintely long wire. We found the magnitude of magnetic field for finite wire was.

The limit \(L/s \rightarrow \infty\) on the right side gives

The direction is given by the right hand rule. As seen in that example, magnetic fields wil go around the wire in circles.

(a) We follow the procedures we have followed in other examples. First, we select an infintesimal element of the wire and then try to set up Biot-Savart expression for infintesimal magnetic field, which we integrate. This is not very efficient method for this problem. For more efficient method, see (b).

Let us place the ring of current in \(xy\) plane as shown in the figure. The elment is a tiny arc of length \(R\,d\phi\text{.}\) The direction of \(d\vec l\) is tangential to the circle. We write this as unit vector.

\begin{equation*} d\vec l = R\,d\phi\, \hat u_t, \end{equation*}

The unit vector in the tangent to the circle direction will be

\begin{equation*} \hat u_t = -\sin\,\phi\, \hat u_x + \cos\,\phi\, \hat j. \end{equation*}

The other vectors \(\vec r\) from infinitesimal element \((R\cos\,\phi, R\sin\,\phi, 0)\) to field point P at \((0, 0, z)\text{,}\) we need for Biot-Savart law are

Now, we have all the parts to write the expression for \(d\vec B_P\text{.}\)

Integrating from \(\phi=0\) to \(\phi=2\pi\text{,}\) only the \(z\) component is nonzero.

This gives the field at P with magnitude \(\dfrac{\mu_0 I}{2}\dfrac{R^2}{ \left(R^2 + z^2\right)^{3/2} }\) in the direction away from the ring for a point above the ring. If we had done for a point below the ring, we would have found that field is pointed towards the ring.

We note several simplifying features of this problem.

1. The problem has a useful symmetry: all points of the ring are at an equal distance from the field point P.
2. Since the direction of the current at any point of the ring is along the tangent to the circle, the angle between the current direction in any element and the direction from the element to the field point P is \(90^{\circ}\text{.}\)

Using these simplifying aspects of the problem, we find that magnetic field from each element \(dl\) of the current loop will have the same magnitude given by

However, the directions of the field at point P from current in different elements of the ring are different. Using the right-hand rule for the direction of magnetic field, we find that the directions from different parts of the ring form a cone at point P as shown in Figure 39.1.13.

Therefore, horizontal components will cancel out and we will be left with only the vertical components of magnetic field. We seek the vertical component of a vector of magnitude dB that makes an angle \(\alpha\) with the vertical (see figure).

where \(\cos\alpha\) is

Hence, the magnetic field at P can be written conceptually as

Substituting for \(dB_P\text{,}\) \(\cos\alpha\text{,}\) and \(dl = Rd\phi\text{,}\) we find that integration variable is the polar angle \(\phi\text{.}\) Since the integrand does not have dependence on \(\phi\text{,}\) the integration over \(\phi\) from \(0\) to \(2\pi\) give the following result for the magnitude of the magnetic field at point P.

By right hand rule on each segment, we find that the fields of the four segments are all in the same direction, the direction of out-of-page. Therefore, they will all add. From symmetry, the net field will be field of one wire. The field of a single segment is for the case of a symmetric point by a finite wire as worked out in Checkpoint 39.1.7. Using the formula obtained there, the magnitude will be

The direction is out-page.

The field will be a superposition of fields from two semi-infinite wires and one half circle. The directions of all fields are into-the-page. Therefore they will all add. To get the individual fields, we appeal to the formulas worked out above in exercises. For half-circular segment we use Checkpoint 39.1.12 with height set to zero.

To get field of semi-infinite wire, we use Checkpoint 39.1.9 with \(\theta_2=0\) and \(\theta_1=\pi/2\text{.}\) Angle \(theta_1\) is \(90^\circ\) since as the far end goes to infinity, the angle will go to this value. Since, the conttributions of the two straight parts add here, we will get

The net field will be