Calculating Magnetic Fields and Forces - AP Physics C: Electricity and Magnetism

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An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

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

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

The magnetic field is calculated using our given values.

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_$0=4\pi*10^{-7}$$\frac{Tm}{A}$$

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

We want to find the current so we solve for .

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

Plug in the values.

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

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

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Relevant equations:

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

$F = I L \times B$

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 or $\frac{R}{2}$ , respectively.

Original

$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}$$

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$

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}$

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

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

A stream of protons moving at velocity $\vec{v}$ 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.

Tap to see back →

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

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

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.

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

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

Tap to see back →

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

The magnetic field is calculated using our given values.

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_$0=4\pi*10^{-7}$$\frac{Tm}{A}$$

Tap to see back →

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

We want to find the current so we solve for .

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

Plug in the values.

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

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

Tap to see back →

Relevant equations:

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

$F = I L \times B$

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 or $\frac{R}{2}$ , respectively.

Original

$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}$$

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$

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}$

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

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

A stream of protons moving at velocity $\vec{v}$ 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.

Tap to see back →

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

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

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.

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

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

Tap to see back →

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

The magnetic field is calculated using our given values.

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_$0=4\pi*10^{-7}$$\frac{Tm}{A}$$

Tap to see back →

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

We want to find the current so we solve for .

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

Plug in the values.

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

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

Tap to see back →

Relevant equations:

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

$F = I L \times B$

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 or $\frac{R}{2}$ , respectively.

Original

$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}$$

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$

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}$

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

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

A stream of protons moving at velocity $\vec{v}$ 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.

Tap to see back →

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

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

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.

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

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

Tap to see back →

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

The magnetic field is calculated using our given values.

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_$0=4\pi*10^{-7}$$\frac{Tm}{A}$$

Tap to see back →

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

We want to find the current so we solve for .

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

Plug in the values.

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

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

Tap to see back →

Relevant equations:

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

$F = I L \times B$

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 or $\frac{R}{2}$ , respectively.

Original

$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}$$

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$

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}$

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

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

A stream of protons moving at velocity $\vec{v}$ 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.

Tap to see back →

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

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

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.

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance away from the wire.

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

Tap to see back →

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

The magnetic field is calculated using our given values.

A solenoid is long and it is made up of turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_$0=4\pi*10^{-7}$$\frac{Tm}{A}$$

Tap to see back →

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

We want to find the current so we solve for .

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

Plug in the values.

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

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

Tap to see back →

Relevant equations:

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

$F = I L \times B$

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 or $\frac{R}{2}$ , respectively.

Original

$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}$$

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$

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}$

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

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

A stream of protons moving at velocity $\vec{v}$ 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.

Tap to see back →

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

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

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.