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AP Physics C Electricity : Calculating Magnetic Fields and Forces

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #11 : Magnetism

An infinitely long wire has a current of $\displaystyle 3A$ running through it. Calculate the magnetic field at a distance 2cm $\displaystyle 2cm$ away from the wire.

$\mu_0=4\pi *10^{-7}\frac{Tm}{A}$ $\displaystyle \mu_0=4\pi *10^{-7}\frac{Tm}{A}$

Possible Answers:

$3*10^{-5}T$ $\displaystyle 3*10^{-5}T$

$2*10^{-4}T$ $\displaystyle 2*10^{-4}T$

$1.5*10^{-3}T$ $\displaystyle 1.5*10^{-3}T$

$9*10^{-5}T$ $\displaystyle 9*10^{-5}T$

$7.4*10^{-7}T$ $\displaystyle 7.4*10^{-7}T$

Correct answer:

$3*10^{-5}T$ $\displaystyle 3*10^{-5}T$

Explanation:

For infinitely long wires, the formula for the magnetic field is $B=\frac{\mu_0I}{2\pi r}$ $\displaystyle B=\frac{\mu_0I}{2\pi r}$ , where $\displaystyle I$ is the current and $\displaystyle r$ is the distance from the wire.

The magnetic field is calculated using our given values.

$B=\frac{(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}})(3\text{A})}{2\pi(0.02\text{m})}$ $\displaystyle B=\frac{(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}})(3\text{A})}{2\pi(0.02\text{m})}$

$B=3*10^{-5}\ \text{T}$ $\displaystyle B=3*10^{-5}\ \text{T}$

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Example Question #193 : Ap Physics C

A solenoid is 50cm $\displaystyle 50cm$ long and it is made up of 800 $\displaystyle 800$ turns of wire. How much current must run through the solenoid to generate a magnetic field of $\displaystyle 1T$ inside of the solenoid?

$\mu_0=4\pi*10^{-7}\frac{Tm}{A}$ $\displaystyle \mu_0=4\pi*10^{-7}\frac{Tm}{A}$

Possible Answers:

$274\ \text{A}$ $\displaystyle 274\ \text{A}$

$97\ \text{A}$ $\displaystyle 97\ \text{A}$

$1237\ \text{A}$ $\displaystyle 1237\ \text{A}$

$583\ \text{A}$ $\displaystyle 583\ \text{A}$

$497\ \text{A}$ $\displaystyle 497\ \text{A}$

Correct answer:

$497\ \text{A}$ $\displaystyle 497\ \text{A}$

Explanation:

The formula for the magnetic field inside the solenoid is $B=\frac{N}{L}\mu_0I$ $\displaystyle B=\frac{N}{L}\mu_0I$ , where $\displaystyle N$ is the number of turns of wire, $\displaystyle L$ is the length of the solenoid, and $\displaystyle I$ is the current.

We want to find the current so we solve for $\displaystyle I$ .

$I=\frac{BL}{N\mu_0}$ $\displaystyle I=\frac{BL}{N\mu_0}$

Plug in the values.

$I=\frac{(1\text{T})(0.5\text{m})}{(800)(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}{}})}$ $\displaystyle I=\frac{(1\text{T})(0.5\text{m})}{(800)(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}{}})}$

$I=497\ \text{A}$ $\displaystyle I=497\ \text{A}$

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Example Question #11 : Magnetism

Two parallel wires a distance $\displaystyle R$ apart each carry a current $\displaystyle I$ , and repel each other with a force $\displaystyle F$ per unit length. If the current in each wire is doubled to $\displaystyle 2I$ , and the distance between them is halved to $\frac{R}{2}$ $\displaystyle \frac{R}{2}$ , by what factor does the force per unit length change?

Possible Answers:

$\displaystyle 2$

$\displaystyle 4$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\displaystyle 8$

Correct answer:

$\displaystyle 8$

Explanation:

Relevant equations:

$\oint B\cdot dl = \mu _o I$ $\displaystyle \oint B\cdot dl = \mu _o I$

$F = I L \times B$ $\displaystyle F = I L \times B$

Step 1: Find the original and new magnetic fields created by wire 1 at wire 2, using Ampere's law with an Amperian loop of radius $\displaystyle R$ or $\frac{R}{2}$ $\displaystyle \frac{R}{2}$ , respectively.

Original

$B (2\pi R) = \mu _o I \rightarrow B = \frac{\mu _o I}{2 \pi R}$ $\displaystyle B (2\pi R) = \mu _o I \rightarrow B = \frac{\mu _o I}{2 \pi R}$

New

$B (2\pi (\frac{R}{2})) = \mu _o (2I) \rightarrow B = \frac{\mu _o (2I)}{2\pi \frac{R}{2}} = \frac{4\mu _o I}{2\pi R}$ $\displaystyle B (2\pi (\frac{R}{2})) = \mu _o (2I) \rightarrow B = \frac{\mu _o (2I)}{2\pi \frac{R}{2}} = \frac{4\mu _o I}{2\pi R}$

Since the wires are parallel to each other and wire 1's field is directed circularly around it, in each case wire 1's field is perpendicular to wire 2.

Step 2: Find the original and new magnetic forces per unit length on wire 2, due to the field created by wire 1.

$F = IL \times B \rightarrow \frac{F}{L} = IB$ $\displaystyle F = IL \times B \rightarrow \frac{F}{L} = IB$

Original

$\frac{F}{L} = I * \frac{\mu _o I }{2 \pi R} = \frac{\mu _o I^2}{2\pi R}\textup{}$

New

$\frac{F}{L} = 2I * \frac{4\mu _o I}{2\pi R} = \frac{8 \mu _o I^2}{2 \pi R}$ $\displaystyle \frac{F}{L} = 2I * \frac{4\mu _o I}{2\pi R} = \frac{8 \mu _o I^2}{2 \pi R}$

So, the new force per unit length is 8 times greater than the original.

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Example Question #11 : Magnetism

A region of uniform magnetic field, $\displaystyle B$ , is represented by the grey area of the box in the diagram. The magnetic field is oriented into the page.

Ps0_movingqinbfield

A stream of protons moving at velocity $\vec{v}$ $\displaystyle \vec{v}$ is directed into the region of the magnetic field, as shown. Identify the correct path of the stream of protons after they enter the region of magnetic field.

Possible Answers:

A semi-circular path oriented horizontally out of the page

A semi-circular path oriented vertically downward

A semi-circular path oriented vertically upward

A semi-circular path oriented horizontally into the page

Correct answer:

A semi-circular path oriented vertically upward

Explanation:

The magnetic force on a moving charged particle is given by the equation:

$\vec{F}=q\,\vec{v}\times \vec{B}$ $\displaystyle \vec{F}=q\,\vec{v}\times \vec{B}$

Isolating the directional component of this equation yields the understanding that the resulting force on a moving charged particle is perpendicular to the plane of the velocity vector and magnetic field vector. Using the right-hand-rule on this cross-product shows that the velocity vector right-crossed into the magnetic field vector into the page yields a magnetic force vector upward on a positive charge. This will result in a semi-circular path oriented vertically upward.

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AP Physics C Electricity : Calculating Magnetic Fields and Forces

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #11 : Magnetism

Example Question #193 : Ap Physics C

Example Question #11 : Magnetism

Example Question #11 : Magnetism

All AP Physics C Electricity Resources

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