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At One Instant A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=(-2.5 \mathrm{mT}) \hat{\mathrm{i}}\). At one instant the velocity of the proton is \(\vec{v}=(2000 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) \((4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{k}}\) and \((\mathrm{b})(4.0 \mathrm{~V} / \mathrm{m}) \mathrm{i}\) ?

Short Answer

Expert verified
(a) The magnitude of the net force is \14.4 \times 10^{-19} \mathrm{N}\ in the \hat{\mathrm{k}}\ direction. (b) The magnitude of the net force is \sqrt{(6.4^2 + 8.0^2)} \times 10^{-19} \mathrm{N}\.

Step by step solution

01

Write down the given information

The magnetic field is \(\vec{B}=(-2.5 \mathrm{mT}) \hat{\mathrm{i}}\). Convert this to Tesla: \(\vec{B}=(-2.5 \times 10^{-3} \mathrm{T}) \hat{\mathrm{i}}\). The velocity of the proton is \(\vec{v}=(2000 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). The charge of the proton is \(e = 1.6 \times 10^{-19} \mathrm{C}\).
02

Recall the formula for the net force on a charged particle in electric and magnetic fields

The net force \( \vec{F} \) on a charged particle in electric \( \vec{E} \) and magnetic fields \( \vec{B} \) is given by the Lorentz force equation: \[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \]
03

Calculate the electric force for case (a)

For part (a), the electric field is \((4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{k}}\). The electric force \( \vec{F}_e \) is given by: \[ \vec{F}_e = q \vec{E} = (1.6 \times 10^{-19} \mathrm{C})(4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{k}} = 6.4 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} \]
04

Calculate the magnetic force for case (a)

The magnetic force \( \vec{F}_b \) is given by: \[ \vec{F}_b = q(\vec{v} \times \vec{B}) = (1.6 \times 10^{-19} \mathrm{C})((2000 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} \times (-2.5 \times 10^{-3} \mathrm{T}) \hat{\mathrm{i}}) \]
05

Compute the cross product for case (a)

Using the right-hand rule, \( \hat{\mathrm{j}} \times \hat{\mathrm{i}} = - \hat{\mathrm{k}}\). So the magnetic force is: \[ \vec{F}_b = (1.6 \times 10^{-19} \mathrm{C})(2000 \mathrm{~m} / \mathrm{s})(-2.5 \times 10^{-3} \mathrm{T})(- \hat{\mathrm{k}}) = 8.0 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} \]
06

Calculate the net force for case (a)

The net force is the sum of the electric force and the magnetic force: \[ \vec{F}_{\text{net}} = \vec{F}_e + \vec{F}_b = 6.4 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} + 8.0 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} = 14.4 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} \]
07

Calculate the electric force for case (b)

For part (b), the electric field is \((4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{i}}\). The electric force is: \[ \vec{F}_e = (1.6 \times 10^{-19} \mathrm{C})(4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{i}} = 6.4 \times 10^{-19} \mathrm{N} \hat{\mathrm{i}} \]
08

Magnetic force for case (b) is the same

Since the velocity \(\vec{v}\) and magnetic field \(\vec{B}\) have not changed, the magnetic force \(\vec{F}_b\) is the same as in case (a): \[ \vec{F}_b = 8.0 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} \]
09

Calculate the net force for case (b)

The net force is the vector sum of the electric force and the magnetic force: \[ \vec{F}_{\text{net}} = \vec{F}_e + \vec{F}_b = 6.4 \times 10^{-19} \mathrm{N} \hat{\mathrm{i}} + 8.0 \times 10^{-19} \mathrm{N} \hat{\mathrm{k}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
An electric field (\textbf{E}) is a physical field produced by electrically charged objects. It affects the behavior of charges within the field. The strength of the electric field is measured in volts per meter (V/m).
When a proton (positive charge) is placed in an electric field, it experiences a force in the direction of the field. The magnitude of this force is given by the formula: \( \textbf{F}_e = q \textbf{E} \)
Here, \textbf{F}_e is the electric force, q is the charge of the proton, and \textbf{E} is the electric field. For a proton, the charge q is approximately \( 1.6 \times 10^{-19} \, C \). Plugging in the values, the electric force can be easily calculated. For example, if \( \textbf{E} = (4.0 \, V/m) \, \textbf{k} \), the force on a proton would be \( \textbf{F}_e = (1.6 \times 10^{-19} \, C)(4.0 \, V/m) \, \textbf{k} = 6.4 \times 10^{-19} \, N \, \textbf{k} \).
Magnetic Field
A magnetic field (\textbf{B}) is a field produced by moving electric charges (like a current) or magnetic materials. The field is measured in teslas (T) or milliteslas (mT). Unlike the electric field, the magnetic field exerts a force on moving charges, which depends on the velocity of the charge and the orientation of the field.
The magnetic force (\textbf{F}_b) exerted on a moving charge in a magnetic field is given by the Lorentz force formula: \( \textbf{F}_b = q (\textbf{v} \times \textbf{B}) \)
Here, \textbf{F}_b is the magnetic force, q is the charge of the proton, \textbf{v} is the velocity of the proton, and \textbf{B} is the magnetic field. Let's consider the proton with velocity \( \textbf{v} = (2000 \, m/s) \, \textbf{j} \) moving through a magnetic field \( \textbf{B} = (-2.5 \, mT) \, \textbf{i} \), which converts to \( -2.5 \times 10^{-3} \, T \).
By calculating the cross product and knowing that \( \textbf{j} \times \textbf{i} = - \textbf{k} \), we find the magnetic force as \( \textbf{F}_b = (1.6 \times 10^{-19} \, C)((2000 \, m/s)(-2.5 \times 10^{-3} \, T)(- \textbf{k})) = 8.0 \times 10^{-19} \, N \, \textbf{k} \).
Proton Velocity
Velocity (\textbf{v}) is a vector quantity that refers to the rate at which an object changes its position. In the case of a proton moving in electric and magnetic fields, its velocity determines the direction and magnitude of the forces acting on it.
For our problem, the proton has a velocity \( \textbf{v} = (2000 \, m/s) \, \textbf{j} \). This means the proton is moving along the y-axis at 2000 meters per second.
The components of the Lorentz force depend on this velocity. For instance, the magnetic force is derived from the cross product of velocity and magnetic field vectors. The electric force depends directly on the velocity's absence or presence along the given axis in the field.
Understanding the velocity vector’s direction helps us apply the right-hand rule to find the direction of the magnetic force and to better grasp the resulting movements or trajectory changes of the proton influenced by both electric and magnetic fields.

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