Question

question 15 (1 point)
a vertical wire carries a current straight up in a region where the magnetic field vector points due north. what is the direction of the resulting force on this current?
down
north
east
west

Explanation:

Step1: Recall the Right-Hand Rule

To determine the direction of the magnetic force on a current - carrying wire, we use the right - hand rule for the cross product \(\vec{F}=I\vec{L}\times\vec{B}\) (where \(\vec{F}\) is the force, \(I\) is the current, \(\vec{L}\) is the length vector of the wire in the direction of the current, and \(\vec{B}\) is the magnetic field). The right - hand rule states that if we point the fingers of our right hand in the direction of \(\vec{L}\) (current direction) and then curl them towards \(\vec{B}\) (magnetic field direction), our thumb points in the direction of \(\vec{F}\).

The current is straight up, so the direction of \(\vec{L}\) (current direction) is upward (let's consider the upward direction as the positive \(z\) - axis for simplicity). The magnetic field \(\vec{B}\) points due north (let's consider the north direction as the positive \(y\) - axis).

Step2: Apply the Right - Hand Rule

Using the right - hand rule for the cross product \(\vec{L}\times\vec{B}\):
- Point your right hand fingers in the direction of the current (upward, along the \(z\) - axis).
- Then, curl your fingers towards the direction of the magnetic field (north, along the \(y\) - axis).
- When we do this, our thumb will point in the direction of the east (along the positive \(x\) - axis). Another way to think about it is using the formula for the cross product in terms of unit vectors: \(\hat{z}\times\hat{y}=-\hat{x}\)? Wait, no, wait. Wait, the standard cross - product of unit vectors: \(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{j}\times\hat{k}=\hat{i}\), \(\hat{k}\times\hat{i}=\hat{j}\) and the anti - commutative property \(\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}\). If the current is in the \(+\hat{z}\) direction (\(\vec{L}\) is in \(+\hat{z}\)) and the magnetic field \(\vec{B}\) is in \(+\hat{y}\) (north), then \(\vec{F}=I\vec{L}\times\vec{B}\), so \(\hat{z}\times\hat{y}=-\hat{x}\)? Wait, no, I think I mixed up the axes. Let's re - define the axes properly. Let's take:
- Let the positive \(y\) - axis be north.
- Let the positive \(x\) - axis be east.
- Let the positive \(z\) - axis be up (current direction).

The cross product \(\vec{L}\times\vec{B}\) (where \(\vec{L}\) is in \(+\hat{z}\) and \(\vec{B}\) is in \(+\hat{y}\)): \(\hat{z}\times\hat{y}=-\hat{x}\)? No, wait, the cross - product formula: \(\hat{z}\times\hat{y}=\hat{z}\times\hat{y}\). Using the right - hand rule: hold your right hand so that your fingers point in the direction of \(\vec{L}\) (up, \(\hat{z}\)) and then bend your fingers towards \(\vec{B}\) (north, \(\hat{y}\)). The direction your thumb points is west? Wait, no, I think I made a mistake in the axis definition. Let's use the right - hand rule correctly:

The right - hand rule for the force on a current - carrying conductor: \(F = ILB\sin\theta\), and the direction is given by the right - hand rule where:
- Point your right hand's index finger in the direction of the current (\(I\)).
- Point your middle finger in the direction of the magnetic field (\(B\)).
- Your thumb will point in the direction of the force (\(F\)).

So current (index finger) is up, magnetic field (middle finger) is north. So when index is up and middle is north, the thumb points to the west? Wait, no, let's do it physically. If the current is going up (out of the page, if we consider the page as the \(xy\) - plane, with north as up on the page, east as right). Wait, maybe a better way:

Imagine the wire is vertical, current going up (towards the sky). The magnetic field is pointing north (towards the north pole of the Earth). To find the force, use the right - hand rule:

- Hold your right hand so that your fingers are in the direction of the current (up: so your thumb is up).
- Then, your fingers should curl towards the magnetic field direction (north). So if your thumb is up (current) and you curl your fingers towards north (magnetic field), the direction of the force (the way your palm pushes) or the direction your fingers would push? Wait, no, the correct right - hand rule for \(\vec{F}=I\vec{l}\times\vec{B}\) is:

The right - hand rule for cross product: align your right hand so that your fingers point in the direction of \(\vec{l}\) (current) and then rotate your hand so that your fingers can curl towards \(\vec{B}\) (magnetic field). The thumb will point in the direction of \(\vec{F}\).

If \(\vec{l}\) is in the \(+\hat{z}\) (up) and \(\vec{B}\) is in \(+\hat{y}\) (north), then \(\vec{l}\times\vec{B}=\hat{z}\times\hat{y}\). Using the determinant formula for cross product:

\[ \hat{z}\times\hat{y}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 0&0&1\\ 0&1&0 \end{vmatrix}=\hat{i}(0\times0 - 1\times1)-\hat{j}(0\times0 - 1\times0)+\hat{k}(0\times1 - 0\times0)=-\hat{i} \]

Wait, \(\hat{i}\) is east? No, \(\hat{i}\) is usually east, \(\hat{j}\) is north, \(\hat{k}\) is up. So \(-\hat{i}\) is west? Wait, no, I think I messed up the sign. Wait, the cross product \(\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}\). So \(\vec{B}\times\vec{l}=\hat{y}\times\hat{z}=\hat{x}\) (east). But \(\vec{F}=I\vec{l}\times\vec{B}\), so \(\vec{l}\times\vec{B}=-\vec{B}\times\vec{l}\), so \(\vec{l}\times\vec{B}=-\hat{x}\) (west)? Wait, now I'm confused. Let's use a different approach. Let's recall the left - hand rule for motors (Fleming's left - hand rule), which is used for the force on a current - carrying conductor in a magnetic field. Fleming's left - hand rule:

- Hold your left hand such that the forefinger points in the direction of the magnetic field (\(B\)).
- The middle finger points in the direction of the current (\(I\)).
- The thumb points in the direction of the force (\(F\)).

So magnetic field (forefinger) - north, current (middle finger) - up (towards the sky). So forefinger north, middle finger up, then the thumb points to the west? Wait, no, if forefinger is north (pointing north), middle finger is up (pointing up), then the thumb points to the west? Wait, no, let's position the hand:

- Forefinger: north (so pointing to the north, along the \(y\) - axis).
- Middle finger: up (along the \(z\) - axis).
- Then the thumb: when we hold the left hand with forefinger north and middle finger up, the thumb points to the west? Wait, no, maybe I had the axes wrong. Let's take the standard coordinate system where:

- \(x\) - axis: east (positive \(x\) is east)
- \(y\) - axis: north (positive \(y\) is north)
- \(z\) - axis: up (positive \(z\) is up)

Fleming's left - hand rule: forefinger (\(B\)) - \(y\) - axis (north), middle finger (\(I\)) - \(z\) - axis (up), then the thumb (\(F\)) will be in the direction of \(-x\) (west)? Wait, no, let's do the cross product correctly. The force on a current - carrying wire is given by \(\vec{F}=I\vec{l}\times\vec{B}\). The vector \(\vec{l}\) is in the direction of the current, so \(\vec{l}=l\hat{z}\) (since current is up), and \(\vec{B}=B\hat{y}\) (since magnetic field is north). Then \(\vec{l}\times\vec{B}=l\hat{z}\times B\hat{y}=lB(\hat{z}\times\hat{y})\). We know that \(\hat{z}\times\hat{y}=-\hat{x}\) (from the cross - product of unit vectors: \(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{j}\times\hat{k}=\hat{i}\), \(\hat{k}\times\hat{i}=\hat{j}\), and \(\hat{z}\) is \(\hat{k}\), \(\hat{y}\) is \(\hat{j}\), so \(\hat{k}\times\hat{j}=-\hat{i}\)). Since \(\hat{i}\) is the east direction, \(-\hat{i}\) is the west direction? Wait, no, \(\hat{i}\) is east, so \(-\hat{i}\) is west. Wait, but earlier when I thought of the right - hand rule, I was confused. Wait, no, Fleming's left - hand rule is for the motor effect, and the right - hand rule for cross product: let's use the right hand, fingers in the direction of \(\vec{l}\) (up, \(\hat{z}\)), then curl towards \(\vec{B}\) (north, \(\hat{y}\)). The thumb will point in the direction of \(\hat{z}\times\hat{y}\). Since \(\hat{z}\times\hat{y}=-\hat{x}\) (west), so the force is in the west direction? Wait, but the options are down, north, east, west. Wait, now I think I made a mistake in the cross - product. Wait, \(\hat{y}\times\hat{z}=\hat{x}\) (east), and since \(\vec{l}\times\vec{B}=-\vec{B}\times\vec{l}\), so \(\vec{l}\times\vec{B}=-\hat{x}\) (west). Wait, but let's check with an example. Suppose the magnetic field is north (towards the top of the page), current is up (out of the page). Then the force on the wire: using the right - hand rule, if you have current out of the page (up) and magnetic field up the page (north), the force is to the left (west) or right (east)? Wait, no, let's use the right - hand rule for the cross product \(\vec{F}=I\vec{l}\times\vec{B}\). The direction of \(\vec{l}\) is the direction of the current (out of the page, \(\hat{z}\)), \(\vec{B}\) is north (up the page, \(\hat{y}\)). So \(\hat{z}\times\hat{y}\): to find this, point your right hand fingers in the direction of \(\hat{z}\) (out of the page, so your thumb is pointing out), then curl your fingers towards \(\hat{y}\) (up the page). The direction your fingers curl is from \(\hat{z}\) to \(\hat{y}\), which is a clockwise rotation in the \(yz\) - plane. The thumb of your right hand, when you do this, will point to the left (west), which is the \(-\hat{x}\) direction. So the force is in the west direction. Wait, but the options include west. Wait, but earlier when I thought of Fleming's left - hand rule, maybe I used the wrong hand. Wait, Fleming's left - hand rule: forefinger (magnetic field), middle finger (current), thumb (force). So magnetic field (forefinger) north, current (middle finger) up, then thumb points west. Yes, that's correct. So the force is in the west direction.

Answer:

west (the option: west)