Problem 60 - University Physics with Modern Physics

Question

An electron is moving in the vicinity of a long, straight wire that lies along the \(x\)-axis. The wire has a constant current of 9.00 A in the \(-x\)-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is \(\vec{v} =\) (5.00 \(\times\) 10\(^4\) m/s)\(\hat{\imath}\) - (3.00 \(\times\) 10\(^4\) m/s)\(\hat{\jmath}\), what is the force that the wire exerts on the electron? Express the force in terms of unit vectors, and calculate its magnitude.

Step-by-Step Solution

Answer

The force is \( \vec{F} = (4.32 \times 10^{-20}\hat{\imath} + 7.20 \times 10^{-20}\hat{\jmath}) \; \text{N} \) with a magnitude of \( 8.41 \times 10^{-20} \; \text{N} \).

Step 1: Identify the Given Values

The wire has a current of \( I = 9.00 \; \text{A} \) in the \(-x\)-direction. At the given instant, the electron is at point \((0, 0.200 \; \text{m}, 0)\) with a velocity of \( \vec{v} = (5.00 \times 10^4 \; \text{m/s})\hat{\imath} - (3.00 \times 10^4 \; \text{m/s})\hat{\jmath} \).

Step 2: Calculate the Magnetic Field Due to the Wire

The magnetic field \( \vec{B} \) around a long, straight wire is given by Ampère's Law: \( B = \frac{\mu_0 I}{2\pi r} \) where \( r \) is the distance from the wire (here, \( r = 0.200 \; \text{m} \)), and \( \mu_0 = 4\pi \times 10^{-7} \; \text{T} \cdot \text{m/A} \). Substitute the values:\[ B = \frac{(4\pi \times 10^{-7} \; \text{T} \cdot \text{m/A})(9.00 \; \text{A})}{2\pi (0.200 \; \text{m})} = 9.00 \times 10^{-6} \; \text{T} \]

Step 3: Determine the Direction of the Magnetic Field

Using the right-hand rule for currents, the magnetic field direction on the specified point (above the wire) is in the negative \(z\)-direction. Thus, \( \vec{B} = -9.00 \times 10^{-6} \; \text{T} \cdot \hat{k} \).

Step 4: Calculate the Magnetic Force on the Electron

The magnetic force \( \vec{F} \) is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \( q = -1.60 \times 10^{-19} \; \text{C} \) is the charge of an electron. Use the cross product:\[ \vec{v} \times \vec{B} = [(5.00 \times 10^4)\hat{\imath} - (3.00 \times 10^4)\hat{\jmath}] \times (-9.00 \times 10^{-6})\hat{k} \]Calculate the cross product component-wise:- \( \hat{\imath} \times \hat{k} = \hat{\jmath} \), so\((-9.00 \times 10^{-6})(5.00 \times 10^4)\hat{\jmath} = -4.5 \times 10^{-1}\hat{\jmath} \)- \( \hat{\jmath} \times \hat{k} = -\hat{\imath} \), so \((-9.00 \times 10^{-6})(-3.00 \times 10^4)(-\hat{\imath}) = -2.7 \times 10^{-1}\hat{\imath} \)Thus, \( \vec{v} \times \vec{B} = -2.7 \times 10^{-1}\hat{\imath} - 4.5 \times 10^{-1}\hat{\jmath} \).

Step 5: Compute the Force

Substitute back into the force formula:\[ \vec{F} = (-1.60 \times 10^{-19} \; \text{C})(-2.7 \times 10^{-1}\hat{\imath} - 4.5 \times 10^{-1}\hat{\jmath}) \] \[ \vec{F} = (4.32 \times 10^{-20}\hat{\imath} + 7.20 \times 10^{-20}\hat{\jmath}) \; \text{N}\]

Step 6: Calculate the Magnitude of the Force

The magnitude of \( \vec{F} \) is \( |\vec{F}| = \sqrt{(4.32 \times 10^{-20})^2 + (7.20 \times 10^{-20})^2} \). Calculate this:\[ |\vec{F}| = \sqrt{(1.87 \times 10^{-39}) + (5.18 \times 10^{-39})} = 8.41 \times 10^{-20} \; \text{N}\]

Key Concepts

Ampère's Law

Ampère's Law is an essential tool for understanding magnetic fields created by electric currents. This fundamental law connects the magnetic field along a closed loop to the electric current flowing through the area embraced by the loop.
The mathematical expression for Ampère's Law is: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I \), where:

• \( \vec{B} \) is the magnetic field.


• \( d\vec{l} \) is a small segment of the loop.


• \( \mu_0 \) is the permeability of free space, with a value of \( 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \).


• \( I \) is the current enclosed by the loop.

In the context of a long, straight wire, Ampère's Law helps us determine the magnetic field at a radial distance \( r \) from the wire. The formula \( B = \frac{\mu_0 I}{2\pi r} \) is derived using a circular path centered on the wire. This expression is particularly useful for calculating magnetic fields in scenarios involving straight, steady currents, such as the wire in our exercise. By knowing the current and distance from the wire, you can easily compute the strength of the magnetic field.

Right-Hand Rule

The right-hand rule is a simple yet effective tool for determining the direction of the magnetic force or magnetic field in a three-dimensional space. It comes into play when dealing with currents or moving charges and their resulting fields. To apply this rule, consider the situation where you have a current-carrying wire and you're interested in the magnetic field around it.
To find the direction of the magnetic field using the right-hand rule for a current-carrying wire:

• Point your right thumb in the direction of the current.


• Encircle the wire with your fingers.
  The direction that your fingers curl represents the direction of the magnetic field lines around the wire.

In the exercise, the current flows along the negative \( x \)-axis. When applying the right-hand rule, you will find that the magnetic field vector at a point ((0, 0.200 \, \text{m}, 0) is directed in the negative z-direction. This understanding of direction is crucial in calculating the overall magnetic force exerted on the moving electron.

Cross Product

The cross product is a mathematical operation that allows you to find a vector perpendicular to two given vectors. It is vital in physics, especially when working with magnetic forces, because the magnetic force vector is always perpendicular to both the velocity vector of a charged particle and the magnetic field vector.
The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is defined as: \( \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin(\theta) \hat{n} \), where:

• \( |\vec{A}| \) and \( |\vec{B}| \) are the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \).


• \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).


• \( \hat{n} \) is the unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \).

In our exercise, the electron's velocity \( \vec{v} \) and the magnetic field \( \vec{B} \) are given, and the cross product \( \vec{v} \times \vec{B} \) is computed to determine the force. The force is calculated using \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \( q \) is the charge of the electron. Through the cross product, you can find each component of the force, which, in turn, helps evaluate the force's direction and magnitude accurately.

Magnetic Field

Magnetic fields are areas where magnetic forces can be detected and have both magnitude and direction. These fields are generated by moving charges, like the flow of electric current in conductors. A magnetic field is represented by the vector \( \vec{B} \), and its effects are observed through the magnetic force it exerts on other charges or magnetic materials.
The strength of the magnetic field created by a long, straight wire is determined by Ampère's Law and is given by the formula \( B = \frac{\mu_0 I}{2\pi r} \). In the exercise, the magnetic field's magnitude is calculated at a specific distance from the wire where the electron is located. The resulting magnetic field influences the force exerted on the electron.
Understanding how magnetic fields interact with moving charges is key in fields like electromechanics and electromagnetism. Magnetic fields follow certain behaviors and rules, such as the right-hand rule, to determine their direction. In this exercise, the magnetic field is directed along the negative z-axis, influencing the force experienced by the electron.