Ch 12 Model Questions

We need the Biot-Savart Law:

\[ \vec{B} = \frac{\mu_0}{4 \pi} q \frac{\vec{v}\times\hat{r}}{r^2} \]

Where \(\hat{r}\) points from the charge to the various numbered points. Remember to reverse the result of the cross product due to the negative charge.

1. \(\vec{B}_1 = \vec{0}\)
   \(\vec{B}_2 = \times\)
   \(\vec{B}_3 = \cdot\)
   \(\vec{B}_4 = \times\)
   \(\vec{B}_5 = \times\)
   \(\vec{B}_6 = \times\)
2. \(B_4 = B_3 > B_6 = B_2 > B_5 > B_1\)

Since point 5 is twice as far as point 4,

\[ B_5 = \tfrac{1}{2^2}B_4 = \tfrac{1}{4} B_4 = 0.25 B_4 \]

Points 2 and 6 are \(\sqrt{2}\) farther away, but include another factor of \(\sin(45^{\circ}) = 1/\sqrt{2}\) (from the cross product). So

\[ B_6 = \tfrac{1}{(\sqrt{2})^2} \tfrac{1}{\sqrt{2}} B_4 \approx 0.3536 \; B_4 > B_5 \]

Repulsion requires south pole on the left side of the loop. So, by the right-hand rule, current flows out of the page on the top part of the loop, and into the page on the lower part of the loop.

The figure shows the net magnetic field (red) and the force (blue) on each current. The light-red vectors are the individual \(\vec{B}\) contributions from each current.

(Note that the force vectors add to zero, as must be the case.)

Another way to find this result is to recall that parallel currents attract and opposite currents repel. Then add the force vectors.

1. Negative. This is required to produce \(B\) field that points outward at point 2, to cancel that of the positive charge.
2. Out of the page. Point 1 is closer to the positive charge.
3. Into of the page. Point 3 is closer to the negative charge.

1. at \((0,0,0)\), \(\vec{B}\) points in the \(-\hat{\jmath}\) direction
2. at \((2,0,0)\), \(\vec{B}\) points in the \(+\hat{\jmath}\) direction
3. at \((0,0,1)\), \(\vec{B}\) points in the \(-\hat{\jmath}\) direction
4. at \((1,1,0)\), \(\vec{B}\) points in the \(-\hat{\imath}\) direction
5. at \((1,0,1)\), \(\vec{B} = \vec{0}\)

The magnetic field is given by

\(\displaystyle \vec{B} = \frac{\mu_0}{4 \pi} \frac{q\vec{v}\times\hat{r}}{r^2}\)

where \(\hat{r}\) points from the particle at \((1,0,0,)\) to the location where we want the field.

For each location, determine the direction of \(\hat{r}\). Then, since \(q\) is positive, \(\vec{B}\) points in the the direction of \(\hat{k}\times\hat{r}\).

1. \(\hat{r} = -\hat{\imath}\), so \(\hat{k}\times(-\hat{\imath}) = -\hat{\jmath}\)
2. \(\hat{r} = +\hat{\imath}\), so \(\hat{k}\times \hat{\imath} = +\hat{\jmath}\)
3. \(\vec{r} = -\hat{\imath} + \hat{k}\), so \(\hat{k}\times( -\hat{\imath} + \hat{k}) = \hat{k}\times (-\hat{\imath}) = -\hat{\jmath}\)
4. \(\hat{r} = +\hat{\jmath}\), so \(\hat{k}\times \hat{\jmath} = -\hat{\imath}\)
5. \(\hat{r} = +\hat{k}\), so \(\hat{k}\times \hat{k} = \vec{0}\)

Note: Another way to get the solutions is to sketch out the coordinates and use the right-hand rule for circulating magnetic field.

• When \(\vec{B}\) and the path are in the same direction, \(\int \vec{B} \cdot d\vec{s}\) is positive.
• When \(\vec{B}\) and the path are in the opposite direction, \(\int \vec{B} \cdot d\vec{s}\) is negative.
• When \(\vec{B}\) and the path are perpendicular \(\int \vec{B} \cdot d\vec{s} = 0\).

Integrals along portions of the loop are

Solutions are

• Loop 1: \(\displaystyle \oint \vec{B} \cdot d \vec{\ell} = 0\)
• Loop 2: \(\displaystyle \oint \vec{B} \cdot d \vec{\ell} = 4\) T·m
• Loop 3: \(\displaystyle \oint \vec{B} \cdot d \vec{\ell} = 0\)

1. \(2 \mu_0\)
2. \(9 \mu_0\)
3. \(0\)
4. \(2 \mu_0\)
5. \(-5 \mu_0\)

Near segment feels an attractive force \(\displaystyle F = B_1 I_2 a\), where \(\displaystyle B_1 = \frac{\mu_0 I_1}{2 \pi \cdot b}\)

Far segment feels a repulsive force \(\displaystyle F = B_2 I_2 a\), where \(\displaystyle B_2 = \frac{\mu_0 I_1}{2 \pi \cdot 2b}\)

Net force is the difference:

\(\displaystyle F = \frac{\mu_0 I_1 I_2 a}{4 \pi b}\) to the left

Use Ampère’s Law, with circles of the given radii.

1. \(0\)
2. \(1.09 \times 10^{-4}\) T, clockwise
3. \(1.67 \times 10^{-4}\) T, clockwise

Use the equation for field in a solenoid: \(\displaystyle B_{\rm{sol}} = \frac{\mu_0 N I}{\ell}\)

Here \(\displaystyle \frac{N}{\ell} = 2000\) turns per meter.

\(B = 1.3 \times 10^{-2}\) T

Solution: \(\displaystyle \vec{B} = \frac{\mu_0 I D \hat{k}}{4 \pi} \int_{x_1}^{x_2} (x^2 + D^2)^{-3/2} dx\)

To obtain this answer, start with the Biot-Savart equation: \(\displaystyle \vec{B} = \int d\vec{B} = \frac{\mu_0}{4 \pi} \int \frac{I d \vec{\ell}\times\hat{r}}{r^2}\).

Since \(I\) is constant, it can come out of the integral. Since we are integrating along the \(x\) axis, \(d\vec{\ell} = \hat{\imath} \; dx\). Also, we can write \(\displaystyle \hat{r} = \frac{\vec{r}}{r}\), so \(\displaystyle \frac{\hat{r}}{r^2} = \frac{\vec{r}}{r^3}\).

Thus

\(\displaystyle \vec{B} = \frac{\mu_0 I}{4 \pi} \int \frac{\hat{\imath}\times\vec{r}}{r^3} dx\)

For a small \(dx\) at location \(x\), \(\vec{r} = x \hat{\imath} + D \hat{\jmath}\) and \(r = \sqrt{x^2 + D^2}\).

For the cross product: \(\hat{\imath}\times\vec{r} = D \hat{k}\).

With these substitutions, the integral becomes

\(\displaystyle \vec{B} = \frac{\mu_0 I}{4 \pi} \int_{x_1}^{x_2} \frac{D \hat{k}}{ \left ( \sqrt{x^2 + D^2} \right )^3} dx\)

which simplifies to the solution above.

To go a bit further, you can solve this by looking up the integral in the textbook, Appendix E, where it says that

\[ \int \frac{1}{(x^2 + a^2)^{3/2}} dx = \frac{x}{a^2 \sqrt{x^2 + a^2}}\]

So

\(\displaystyle \vec{B} = \frac{\mu_0 I D \hat{k}}{4 \pi} \left [ \frac{x}{D^2 \sqrt{x^2 + D^2}} \right ]_{x_1}^{x_2} = \frac{\mu_0 I \hat{k}}{4 \pi D} \left [ \frac{x_2}{\sqrt{x_2^2+D^2}} - \frac{x_1}{\sqrt{x_1^2+D^2}} \right ]\)