Understanding Magnetic Fields and Wires - AP Physics C Electricity & Magnetism

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Question

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

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 ?

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

Answer

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

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

$B_{1}=\left| \frac{(4\pi10^{-7}\frac{T\cdot m}{A})(6A)}{2\pi (1.5mm)} \right|=810^{-4}T$

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

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

$B_{2}=\left| \frac{(4\pi10^{-7}\frac{T\cdot m}{A})(9A)}{2\pi (3.0mm)} \right|=610^{-4}T$

The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude $1.410^{-3}T$ to the right.

$810^{-4}T+610^{-4}T=1.410^{-3}T$

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Question

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

What is the direction of the magnetic field produced?

Answer

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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Question

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

What is the magnitude of the magnetic field at point ?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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

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 ?

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

Answer

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

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

$B_{1}=\left| \frac{(4\pi10^{-7}\frac{T\cdot m}{A})(6A)}{2\pi (1.5mm)} \right|=810^{-4}T$

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

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

$B_{2}=\left| \frac{(4\pi10^{-7}\frac{T\cdot m}{A})(9A)}{2\pi (3.0mm)} \right|=610^{-4}T$

The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude $1.410^{-3}T$ to the right.

$810^{-4}T+610^{-4}T=1.410^{-3}T$

Compare your answer with the correct one above

Question

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

What is the direction of the magnetic field produced?

Answer

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.

Compare your answer with the correct one above

Question

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

What is the magnitude of the magnetic field at point ?

Answer

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:

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

Compare your answer with the correct one above

Question

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?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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:

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

Compare your answer with the correct one above

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