Question

A particle with a charge of -1.24 $\times$ 10${}^{-8}$ C is moving with instantaneous velocity $\overrightarrow{v}=$ 14.19 $\times$ 10${}^4$ m/s)$\hat{ı}$ + (-3.85 $\times$ 10${}^4$ m/s)$\hat{ȷ}$. What is the force exerted on this particle by a magnetic field (a) $\overrightarrow{B}=$ (1.40 T)$\hat{ı}$ and (b) $\overrightarrow{B}=$ (1.40 T) $\hat{k}$ ?

Short answer

(a) $\overrightarrow{F_1}=0\hat{ı}+0\hat{ȷ}-6.68\times {10}^{-4}\hat{k}\text{ N}$; (b) $\overrightarrow{F_2}=6.68\times {10}^{-4}\hat{ı}+2.47\times {10}^{-3}\hat{ȷ}\text{ N}$.

Step-by-step solution

Understand the Given Problem

We need to find the force exerted by a magnetic field on a charge with a specific velocity vector in two different scenarios. The charge is given as $q=-1.24\times {10}^{-8}\text{ C}$, and the velocity vector is $\overrightarrow{v}=14.19\times {10}^4\hat{ı}-3.85\times {10}^4\hat{ȷ}\text{ m/s}$. We need to find the force for two different magnetic fields: $\overrightarrow{B_1}=1.40\hat{ı}\text{ T}$ and $\overrightarrow{B_2}=1.40\hat{k}\text{ T}$.

Use the Lorentz Force Formula

The magnetic force $\overrightarrow{F}$ on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the cross product $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. We'll need to compute this for each magnetic field scenario.

Calculate Cross Product for (a) $\overrightarrow{B}=1.40\hat{ı}$

For $\overrightarrow{B_1}=1.40\hat{ı}$, the cross product is $\overrightarrow{v}\times \overrightarrow{B_1}=(14.19\times {10}^4\hat{ı}-3.85\times {10}^4\hat{ȷ})\times 1.40\hat{ı}$. Since a vector crossed with itself is zero, and the cross product between vectors on the same axis is zero, this simplifies to: $\overrightarrow{v}\times \overrightarrow{B_1}=0\hat{ı}+0\hat{ȷ}+(-1.40\times (-3.85\times {10}^4))\hat{k}$. Calculating gives: $\overrightarrow{v}\times \overrightarrow{B_1}=0\hat{ı}+0\hat{ȷ}+5.39\times {10}^4\hat{k}$.

Calculate Force for (a)

Using $q=-1.24\times {10}^{-8}\text{ C}$ and the cross product from Step 3, we find: $\overrightarrow{F_1}=-1.24\times {10}^{-8}\times (0\hat{ı}+0\hat{ȷ}+5.39\times {10}^4\hat{k})$. This gives: $\overrightarrow{F_1}=0\hat{ı}+0\hat{ȷ}-6.68\times {10}^{-4}\hat{k}\text{ N}$.

Calculate Cross Product for (b) $\overrightarrow{B}=1.40\hat{k}$

For $\overrightarrow{B_2}=1.40\hat{k}$, the cross product is $\overrightarrow{v}\times \overrightarrow{B_2}=(14.19\times {10}^4\hat{ı}-3.85\times {10}^4\hat{ȷ})\times 1.40\hat{k}$.The result is calculated using the determinant method for cross products:$(-3.85\times {10}^4\times 1.40)\hat{ı}-(14.19\times {10}^4\times 1.40)\hat{ȷ}$. Which results in:$\overrightarrow{v}\times \overrightarrow{B_2}=-5.39\times {10}^4\hat{ı}-1.99\times {10}^5\hat{ȷ}+0\hat{k}$.

Calculate Force for (b)

Using $q=-1.24\times {10}^{-8}\text{ C}$ and the cross product from Step 5, we find:$\overrightarrow{F_2}=-1.24\times {10}^{-8}(-5.39\times {10}^4\hat{ı}-1.99\times {10}^5\hat{ȷ})$.This results in: $\overrightarrow{F_2}=6.68\times {10}^{-4}\hat{ı}+2.47\times {10}^{-3}\hat{ȷ}\text{ N}$.

Key concepts

Magnetic Field

A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It's visualized as "lines of force" that exit through a magnet's north pole and enter through its south pole. Magnetic fields can be created by permanent magnets or flows of electric current. When a moving charge enters a magnetic field, it experiences a force. This force's direction depends on the charge's velocity and the magnetic field's direction. The strength of a magnetic field is measured in Tesla (T).

Key Points:

• A magnetic field exerts a force on moving charges.
• The magnetic field's direction and the charge’s motion will determine the direction of this force.
• Strong magnetic fields exert stronger forces on charges.

In the context of the original exercise, recognizing which direction the magnetic fields ($\overrightarrow{B_1}$ and $\overrightarrow{B_2}$) are acting in is crucial for calculating the resulting force on the charged particle.

Cross Product

The cross product is a vector multiplication operation that yields a vector perpendicular to two input vectors in three-dimensional space. This operation is central in physics, particularly in electromagnetism, where it is used to determine the force experienced by a charged particle within a magnetic field. Mathematically, the cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is denoted as $\overrightarrow{a}\times \overrightarrow{b}$. The magnitude of the resulting vector is given by $|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$, where $\theta$ is the angle between the original vectors.
Key Characteristics:

• A cross product of parallel vectors is zero since $\sin (0^{\circ })=0$.
• The resulting vector is perpendicular to the original vectors.
• It is used to calculate torques, rotations, and forces like the Lorentz force.

In the exercise, the cross product helps determine $\overrightarrow{v}\times \overrightarrow{B}$, which then allows calculation of the Lorentz force on the particle through $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$.

Vector Calculus

Vector calculus is a mathematical tool used to study vector fields, which are functions assigning a vector to each point in space. It's vital in physics for understanding phenomena involving fields, such as electromagnetic fields. Vector operations, such as addition, scalar multiplication, and cross products, help solve complex physical problems.

Important Operations:

• Vector Addition: Combines two vectors' magnitudes and directions.
• Scalar Multiplication: Involves multiplying a vector by a scalar (number), scaling its magnitude.
• Cross Product: Yields a vector perpendicular to two given vectors, especially crucial in calculating forces in electromagnetism.

In our task, vector calculus is applied to the velocity vector $\overrightarrow{v}$ and the magnetic field $\overrightarrow{B}$ to determine the force using the cross product. This application exemplifies the real-world utility of vector calculus in physics.