Problem 27 - University Physics with Modern Physics

Question

A particle with initial velocity \(\vec{v}$$_0 =\) (5.85 \(\times\) 10\(^3\)m/s)\(\hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{B} =\) - (1.35 T)\(\hat{k}\). Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

Step-by-Step Solution

Answer

The electric field is \((-7.8975 \times 10^3 \, \text{V/m}) \hat{\imath}\) for both charges.

Step 1: Understanding the condition for no deflection

For a charged particle to pass through a region with electric and magnetic fields without being deflected, the electric force and magnetic force must cancel each other out. This can be expressed as \( \vec{F}_e + \vec{F}_m = 0 \), where \( \vec{F}_e = q \vec{E} \) and \( \vec{F}_m = q(\vec{v} \times \vec{B}) \). Therefore, \( \vec{E} = -\vec{v} \times \vec{B} \).

Step 2: Calculating the cross product \(\vec{v} \times \vec{B}\)

Given \( \vec{v}_0 = (5.85 \times 10^3 \, \text{m/s})\hat{\jmath} \) and \( \overrightarrow{B} = - (1.35 \, \text{T})\hat{k} \), the cross product is calculated by: \[ \vec{v} \times \vec{B} = (5.85 \times 10^3 \hat{\jmath}) \times (-1.35 \hat{k}) = -1.35 \times 5.85 \times 10^3 \hat{\imath} \].Using the right-hand rule, \( \hat{\jmath} \times (-\hat{k}) = \hat{\imath} \), therefore,\[ \vec{v} \times \vec{B} = (7.8975 \times 10^3)\hat{\imath} \].

Step 3: Calculate the electric field required

Since \( \vec{E} = -\vec{v} \times \vec{B} \), we find that:\[ \vec{E} = -(7.8975 \times 10^3 \hat{\imath}) = (-7.8975 \times 10^3) \hat{\imath} \].This shows that the electric field must be \((-7.8975 \times 10^3 \, \text{V/m}) \hat{\imath}\) to ensure the particle remains undeflected.

Step 4: Solution Interpretation for both charges

The result is independent of the charge magnitude and sign because it isn't relevant when calculating the required electric field for zero net force:(a) For the particle with charge +0.640 nC, the electric field direction \( (-\hat{\imath}) \) means it is in the negative x-direction.(b) For the particle with charge -0.320 nC, the electric field remains \((-7.8975 \times 10^3 \, \text{V/m}) \hat{\imath}\), as within the cross product setup, the charge affects force magnitude but not the field needed to counter magnetic effects.

Key Concepts

Electric Field

An electric field is a region around a charged particle where a force would be exerted on other charges. It is a vector quantity, which means it has both magnitude and direction.
The electric field, represented as \( \vec{E} \), influences charged particles by exerting a force on them, given by the equation \( \vec{F}_e = q \vec{E} \), where \( q \) is the charge of the particle.

• The direction of the electric field is the direction of the force it exerts on a positive charge.
• Electric fields are often depicted using lines, called field lines, originating from positive charges and terminating at negative charges.

In the context of the problem, the electric field is used to counteract the force exerted on the particle by the magnetic field. This keeps the particle moving in a straight line, undeflected.

Magnetic Field

A magnetic field is a field produced by moving electric charges and magnetic dipoles, and it exerts forces on other moving charges. Like electric fields, magnetic fields are also vector fields, denoted by \( \overrightarrow{B} \).
The magnetic force on a moving charge can be computed with the formula \( \vec{F}_m = q(\vec{v} \times \vec{B}) \). This shows that the force is perpendicular to both the velocity \( \vec{v} \) of the particle and the magnetic field \( \overrightarrow{B} \).

• Magnetic field lines form closed loops and can be visualized as emerging from the north pole of a magnet and entering a south pole.
• The strength of a magnetic field is measured in teslas (T).

In the problem, the provided magnetic field is \( -1.35 \hat{k} \) Tesla, affecting the particle's trajectory due to its perpendicular orientation with the velocity.

Charged Particles

Charged particles are particles with an electrical charge, either positive or negative. The electric and magnetic fields apply forces to these particles.
The motion of charged particles under these fields is crucial in many applications, such as cyclotrons in particle physics or deflection in cathode ray tubes.

• Positive charges move in the direction of electric fields, while negative charges move in the opposite direction.
• The overall impact depends on both the sign and magnitude of the charge, alongside the influence of the fields.

Within the exercise, whether the particle has a positive or negative charge, the principle for the balance between the electric and magnetic force remains the same. However, the direction of those forces does depend on the charge's sign, ensuring the path remains undeflected by adjusting the electric field accordingly.

Vector Cross Product

The vector cross product, denoted as \( \vec{A} \times \vec{B} \), is a mathematical operation that takes two vectors and produces a third vector perpendicular to the plane formed by the initial vectors. This operation is crucial in understanding forces in electromagnetic fields.
In mathematical terms, the magnitude of the cross product is given by \( |\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin(\theta) \), where \( \theta \) is the angle between the two vectors. The direction is found using the right-hand rule.

• The thumb of the right hand points along the first vector, fingers point along the second vector, and the perpendicular direction given by the palm is the direction of the cross product vector.
• In magnetic force calculations, the use of cross product ensures the resulting force is perpendicular to both velocity and magnetic field vectors.

In the problem's solution, \( \vec{v} \times \overrightarrow{B} \) is calculated to find the necessary orientation and strength of the electric field to achieve zero net force on the particle.