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AP Physics C Electricity : Understanding Magnetic Fields and Wires

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

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Example Questions

Example Question #191 : Ap Physics C

Two infinitely long wires having currents I_1 and I_2 are separated by a distance .

Ps0_twowiresbfield

The current I_1 is 6A into the page. The current I_2 is 9A into the page. The distance of separation is 1.5mm. The point lies 1.5mm away from I_1 on a line connecting the centers of the two wires.

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

At point , the magnetic field due to I_1 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 I_2 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.

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

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Example Question #2 : Magnetism

Ring

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

What is the direction of the magnetic field produced?

Possible Answers:

Into the screen

There is no resultant magnetic field

Counterclockwise

Out of the screen

Clockwise

Correct answer:

Into the screen

Explanation:

The correct answer is into the page. As the current is moving clockwise, we can use our right hand rule for magnetic fields produced by a current-carrying loop. Curl the fingers of your right hand in the direction of the current. This should result in your thumb pointing toward the screen, indicating the direction of the magnetic field.

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Example Question #1 : Magnetism

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?

Possible Answers:

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 four times weaker than the original

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

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

Correct answer:

The new magnetic field would be eight times as strong as 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.

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Example Question #2 : Magnetism

Ring

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

What is the magnitude of the magnetic field at point ?

Possible Answers:

$\frac{\mu_0I}{2\pi R}$

$\frac{\mu_0I}{2R}$

There is no magnetic field at point

$\frac{\mu_{0}I}{R}$

$\frac{\mu_{0}I}{\pi R^2}$

Correct answer:

$\frac{\mu_0I}{2\pi R}$

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:

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

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Example Question #3 : Magnetism

Consider two long, straight, current-carrying wires at distance from each other, each with a current of magnitude going in opposite directions.

What is the magnitude of the magnetic field at a point equidistant from both wires?

Possible Answers:

$\frac{\mu_0I}{\pi d}$

Zero

$\frac{2\mu_0I}{\pi d}$

$\frac{4\mu_0I}{\pi d}$

$\frac{\mu_0I}{2\pi d}$

Correct answer:

$\frac{2\mu_0I}{\pi d}$

Explanation:

Using our right hand rule for magnetic fields produced by current-carrying wires, we know that the magnetic field produced by each wire is in the same direction within the distance between the wires. Therefore, we know that the total magnetic field is simply the magnetic field of one of the wires multiplied by two.

Use the equation for magnetic field:

$B=\frac{\mu_0I}{2\pi R}$

Multiply by two, since the magnetic field will be equal for each wire, and substitute the given variables:

$B=\frac{2\mu_0I}{2\pi (\frac{1}{2}d)}=\frac{2\mu_0I}{\pi d}$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Understanding Magnetic Fields and Wires

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

All AP Physics C Electricity Resources

Example Questions

Example Question #191 : Ap Physics C

Example Question #2 : Magnetism

Example Question #1 : Magnetism

Example Question #2 : Magnetism

Example Question #3 : Magnetism

All AP Physics C Electricity Resources

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