University

Physics 2

Physics 2

Fall 2024

A negative charge moves downward as shown.

- At each of the six numbered points, sketch the direction of the magnetic field vector. Use \((\times)\) for into the page, \((\cdot)\) for out of the page, and \(\vec{0}\) for zero field.
- Rank in order from strongest to weakest, the magnetic field strengths \(B_1\) to \(B_6\) at these points.

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.

- \(\vec{B}_1 = \vec{0}\)

\(\vec{B}_2 = \times\)

\(\vec{B}_3 = \cdot\)

\(\vec{B}_4 = \times\)

\(\vec{B}_5 = \times\)

\(\vec{B}_6 = \times\) - \(B_4 = B_3 > B_6 = B_2 > B_5 > B_1\)

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 \(\sin(45^{\circ}) = 1/\sqrt{2}\). So

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

The current loop (perpendicular to the page) exerts a repulsive force on the bar magnet. What is the direction of the current in the loop?

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 three current-carrying wires shown are perpendicular to the page. What is the direction of the net force on each wire?

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.

Two charges are moving as shown. At this instant of time, the net magnetic field at point 2 is \(\vec{B}_2 = \vec{0}\).

- Is the unlabeled moving charge positive or negative?
- What is the direction of the magnetic field at point 1?
- What is the direction of the magnetic field at point 3?

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

A proton at location \(\vec{r} =\) (\(1\) m) \(\hat{\imath}\) has a velocity of \(\vec{v} =\) (\(1\) m/s) \(\hat{k}\).

What is the direction of its magnetic field at the following points?

- at the origin
- at coordinates \((2,0,0)\)
- at coordinates \((0,0,1)\)
- at coordinates \((1,1,0)\)
- at coordinates \((1,0,1)\)

Give answer as a unit vector, or as \(\vec{0}\).

- at \((0,0,0)\), \(\vec{B}\) points in the \(-\hat{\jmath}\) direction
- at \((2,0,0)\), \(\vec{B}\) points in the \(+\hat{\jmath}\) direction
- at \((0,0,1)\), \(\vec{B}\) points in the \(-\hat{\jmath}\) direction
- at \((1,1,0)\), \(\vec{B}\) points in the \(-\hat{\imath}\) direction
- 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 \(\hat{k}\times\hat{r}\).

- \(\hat{r} = -\hat{\imath}\), so \(\hat{k}\times(-\hat{\imath}) = -\hat{\jmath}\)
- \(\hat{r} = +\hat{\imath}\), so \(\hat{k}\times \hat{\imath} = +\hat{\jmath}\)
- \(\vec{r} = -\hat{\imath} + \hat{k}\), so \(\hat{k}\times( -\hat{\imath} + \hat{k}) = \hat{k}\times (-\hat{\imath}) = -\hat{\jmath}\)
- \(\hat{r} = +\hat{\jmath}\), so \(\hat{k}\times \hat{\jmath} = -\hat{\imath}\)
- \(\hat{r} = +\hat{k}\), 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.

The magnetic field above the dotted line is \(\vec{B} = 2 \hat{\imath}\) T. Below the dotted line the field is \(\vec{B} =-2 \hat{\imath}\) T. Each closed loop is 1 m × 1 m. Evaluate the line integral \(\displaystyle \oint \vec{B} \cdot d \vec{\ell}\) in a clockwise direction around each loop.

- 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

path | Loop 1 | Loop 2 | Loop 3 |
---|---|---|---|

up the left edge | 0 | 0 | 0 |

rightward across top | \(+2\) | \(+2\) | \(-2\) |

down the right edge | 0 | 0 | 0 |

leftward across bottom | \(-2\) | \(+2\) | \(+2\) |

all around loop | 0 | \(+4\) | 0 |

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: \(\displaystyle \oint \vec{B} \cdot d \vec{\ell} = 0\)

Evaluate \(\displaystyle \oint \vec{B} \cdot d \vec{\ell}\) for each of the cases shown. Give your answer as a factor times \(\mu_0\).

- \(2 \mu_0\)
- \(9 \mu_0\)
- \(0\)
- \(2 \mu_0\)
- \(-5 \mu_0\)

The infinite, straight wire shown carries a current \(I_1\). The rectangular loop, whose long sides are parallel to the wire, carries a current \(I_2\).

In terms of the vaiables given, what is the magnitude and direction of the net force on the rectangular loop due to the magnetic field of the wire?

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

The figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius \(r_1=3.0\) cm and outer radius \(r_2=5.0\) cm. A \(50\) A current distributed uniformly over the cross-section flows into the page.

Calculate the magnetic field (magnitude and direction) at the following radii.

- \(r=2.0\) cm
- \(r=4.0\) cm
- \(r=6.0\) cm

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

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

A solenoid is wound with \(2000\) turns per meter. When the current is \(5.2\) A, what is the magnetic field within the solenoid?

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

Determine the magnetic field at point \(P\) due to the wire segment carrying current \(I\) between point \(x_1\) and \(x_2\).

Set up and simplify the integral required to solve this problem – you do not need to evaluate the integral.

A fully simplified answer should

- be written in terms of the variables shown (and fundamental constants)
- be an integral (with limits) over a single variable
- have all constants out of the integral
- include \(\hat{\imath} \; \hat{\jmath} \; \hat{k}\) (as necessary) for the vector direction

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

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

*Last modified: Wed October 09 2024, 06:23 PM.*