Magnetism

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AP Physics C: Electricity and Magnetism › Magnetism

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Ring

Consider a current-carrying loop with current , radius , and center .

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times weaker than the original

Explanation

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

Ring

Consider a current-carrying loop with current , radius , and center .

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times weaker than the original

Explanation

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

Two infinitely long wires having currents and are separated by a distance .

Ps0_twowiresbfield

The current is 6A into the page. The current is 9A into the page. The distance of separation is 1.5mm. The point lies 1.5mm away from on a line connecting the centers of the two wires.

What is the magnitude and direction of the net magnetic field at the point ?

$\mu_0=4\pi*10^{-7}\frac{T\cdot m}{A}$

$1.4*10^{-3}T\ \text{to the right}$

$1.4*10^{-3}T\ \text{to the left}$

$2.0*10^{-4}T\ \text{to the right}$

$2.0*10^{-3}T\ \text{to the left}$

Explanation

At point , the magnetic field due to points right (via the right hand rule) with a magnitude given by:

$B_{1}=\left| \frac{\mu_{o} I_{1}}{2\pi r_{1}} \right|$

$B_{1}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(6A)}{2\pi (1.5mm)} \right|=8*10^{-4}T$

At point , the magnetic field due to points right (via the right hand rule) with a magnitude given by:

$B_{2}=\left| \frac{\mu_{o} I_{2}}{2\pi r_{2}} \right|$

$B_{2}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(9A)}{2\pi (3.0mm)} \right|=6*10^{-4}T$

The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude $1.4*10^{-3}T$ to the right.

$8*10^{-4}T+6*10^{-4}T=1.4*10^{-3}T$

Two infinitely long wires having currents and are separated by a distance .

Ps0_twowiresbfield

The current is 6A into the page. The current is 9A into the page. The distance of separation is 1.5mm. The point lies 1.5mm away from on a line connecting the centers of the two wires.

What is the magnitude and direction of the net magnetic field at the point ?

$\mu_0=4\pi*10^{-7}\frac{T\cdot m}{A}$

$1.4*10^{-3}T\ \text{to the right}$

$1.4*10^{-3}T\ \text{to the left}$

$2.0*10^{-4}T\ \text{to the right}$

$2.0*10^{-3}T\ \text{to the left}$

Explanation

At point , the magnetic field due to points right (via the right hand rule) with a magnitude given by:

$B_{1}=\left| \frac{\mu_{o} I_{1}}{2\pi r_{1}} \right|$

$B_{1}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(6A)}{2\pi (1.5mm)} \right|=8*10^{-4}T$

At point , the magnetic field due to points right (via the right hand rule) with a magnitude given by:

$B_{2}=\left| \frac{\mu_{o} I_{2}}{2\pi r_{2}} \right|$

$B_{2}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(9A)}{2\pi (3.0mm)} \right|=6*10^{-4}T$

The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude $1.4*10^{-3}T$ to the right.

$8*10^{-4}T+6*10^{-4}T=1.4*10^{-3}T$

Ring

Consider a current-carrying loop with current , radius , and center .

A particle with charge flies through the center and into the page with velocity . What is the total electromagnetic force on the particle at the instant that it flies through the loop, in terms of the variables given?

$QvB\ \text{to the left}$

$QvB\ \text{into the screen}$

$QvB\ \text{out of the screen}$

$QvB\ \text{to the right}$

Explanation

The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.

$\vec F=Q\vec v\times \vec B$

Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.

Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?

Zero regardless of the orientation of the magnet

Positive only if the north pole of the magnet is within the surface

Negative only if the north pole of the magnet is within the surface

Zero only if the magnet is completely enclosed within the surface

More than one of the other options is true

Explanation

The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."

Ring

Consider a current-carrying loop with current , radius , and center .

$QvB\ \text{to the left}$

$QvB\ \text{into the screen}$

$QvB\ \text{out of the screen}$

$QvB\ \text{to the right}$

Explanation

$\vec F=Q\vec v\times \vec B$

Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.

Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?

Zero regardless of the orientation of the magnet

Positive only if the north pole of the magnet is within the surface

Negative only if the north pole of the magnet is within the surface

Zero only if the magnet is completely enclosed within the surface

More than one of the other options is true

Explanation

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?

$\frac{1}{2}$

$\frac{1}{4}$

Explanation

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.

Ps0_movingqinbfield

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.

A semi-circular path oriented vertically upward

A semi-circular path oriented horizontally into the page

A semi-circular path oriented vertically downward

A semi-circular path oriented horizontally out of the page

Explanation

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.

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