5. (a) A horizontal power line of length 65.0 m carries a current of 2.50 kA to the right. The Earth's magnetic field at this location has a magnitude of 50.0 µT. The angle between... 5. (a) A horizontal power line of length 65.0 m carries a current of 2.50 kA to the right. The Earth's magnetic field at this location has a magnitude of 50.0 µT. The angle between the current and the field at this location is 50.0° as shown. (i) Write the magnetic field vector in unit vector notation. (ii) Find the magnetic force vector on the power line in unit vector notation. (b) Four long, parallel conductors carry equal currents of I = 4.00 A. The figure at right shows the cross-sectional view of the four conductors. Determine the magnitude and direction of the net magnetic field at point Q. The length of l is 0.250 m.
Understand the Problem
The problem consists of two parts: (a) involves calculating the magnetic field vector and magnetic force vector on a horizontal power line due to the Earth's magnetic field. Part (b) requires determining the net magnetic field at a point Q due to four long, parallel conductors carrying equal currents.
Answer
(a)(i) $\vec{B} = (-3.21 \times 10^{-5} \, \text{T}) \hat{i} + (3.83 \times 10^{-5} \, \text{T}) \hat{j}$ (a)(ii) $\vec{F} = (6.21 \, \text{N}) \hat{k}$ (b) $B_{net} = 0 \, \text{T}$
Answer for screen readers
(a)(i) $\vec{B} = (-3.21 \times 10^{-5} , \text{T}) \hat{i} + (3.83 \times 10^{-5} , \text{T}) \hat{j}$ (a)(ii) $\vec{F} = (6.21 , \text{N}) \hat{k}$ (b) $B_{net} = 0 , \text{T}$
Steps to Solve
- (a)(i) Find the components of the magnetic field vector
Given that the magnitude of the magnetic field $B = 50.0 , \mu\text{T} = 50.0 \times 10^{-6} , \text{T}$ and the angle between the current and the field is $50.0^\circ$, we can find the components of the magnetic field vector in the $i$ and $j$ directions. The magnetic field vector $\vec{B}$ makes an angle of $50.0^\circ$ with the negative $i$-axis, meaning the angle with the positive $i$-axis is $180^\circ - 50^\circ = 130^\circ$.
$$B_x = B \cos(130^\circ) = (50.0 \times 10^{-6} , \text{T}) \cos(130^\circ) \approx -3.21 \times 10^{-5} , \text{T}$$ $$B_y = B \sin(130^\circ) = (50.0 \times 10^{-6} , \text{T}) \sin(130^\circ) \approx 3.83 \times 10^{-5} , \text{T}$$ Therefore, the magnetic field vector is: $$\vec{B} = (-3.21 \times 10^{-5} , \text{T})\hat{i} + (3.83 \times 10^{-5} , \text{T})\hat{j}$$
- (a)(ii) Calculate the magnetic force vector
The magnetic force on a current-carrying wire is given by $\vec{F} = I \vec{L} \times \vec{B}$, where $I$ is the current, $\vec{L}$ is the length vector, and $\vec{B}$ is the magnetic field vector. Given $I = 2.50 , \text{kA} = 2500 , \text{A}$ and $L = 65.0 , \text{m}$ along the positive $i$-axis, $\vec{L} = (65.0 , \text{m})\hat{i}$. We already found $\vec{B} = (-3.21 \times 10^{-5} , \text{T})\hat{i} + (3.83 \times 10^{-5} , \text{T})\hat{j}$. $$\vec{F} = I \vec{L} \times \vec{B} = (2500 , \text{A}) (65.0 , \text{m} , \hat{i}) \times ((-3.21 \times 10^{-5} , \text{T})\hat{i} + (3.83 \times 10^{-5} , \text{T})\hat{j})$$
We can compute the cross product: $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$ $$\vec{F} = (2500 , \text{A}) (65.0 , \text{m}) (3.83 \times 10^{-5} , \text{T}) \hat{k} = 6.21 , \text{N} , \hat{k}$$ Thus, the magnetic force vector is: $$\vec{F} = (6.21 , \text{N}) \hat{k}$$
- (b) Calculate the magnetic field due to each wire
The magnetic field due to a long, straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$, where $\mu_0 = 4 \pi \times 10^{-7} , \text{T} \cdot \text{m/A}$ is the permeability of free space, $I$ is the current, and $r$ is the distance from the wire to point Q. In this case, $I = 4.00 , \text{A}$ and $r = \frac{l}{2} = \frac{0.250 , \text{m}}{2} = 0.125 , \text{m}$. The magnetic field due to each wire is: $$B = \frac{(4 \pi \times 10^{-7} , \text{T} \cdot \text{m/A}) (4.00 , \text{A})}{2 \pi (0.125 , \text{m})} = \frac{2 \times 10^{-7} \times 4}{0.125} = 6.4 \times 10^{-6} , \text{T}$$
- (b) Determine the direction of the magnetic field due to each wire
The two wires with current going into the page (denoted by X) produce magnetic fields pointing downwards (towards the bottom wire) and to the right (towards the right wire) respectively. The two wires with current coming out of the page (denoted by a dot) produce magnetic fields pointing upwards (towards the top wire) and to the left (towards to the left wire).
- (b) Compute the net magnetic field
Since the magnetic fields due to each pair of opposing wires are in opposite directions they add up. We have two magnetic fields pointing to the right/left and two pointing up/down.
The net magnetic field due to the two wires on the left and right is zero because the separation and magnitude is the same $$ B_{net, x} = B - B = 0 $$ The net magnetic field due to the two wires on the top and bottom is zero because the separation and magnitude is the same. $$ B_{net, y} = B - B = 0 $$ The net magnetic field is therefore zero.
(a)(i) $\vec{B} = (-3.21 \times 10^{-5} , \text{T}) \hat{i} + (3.83 \times 10^{-5} , \text{T}) \hat{j}$ (a)(ii) $\vec{F} = (6.21 , \text{N}) \hat{k}$ (b) $B_{net} = 0 , \text{T}$
More Information
The direction of the magnetic force on the power line is along the positive z-axis (out of the page). In part (b), the symmetry of the problem causes the magnetic fields to cancel at point Q.
Tips
A common mistake in part (a) is to not correctly resolve the magnetic field into its components. In part (b), a common mistake is to not consider the direction of the magnetic field due to each wire and incorrectly add the magnitudes without considering the vector nature of the field.
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