Animated Solution for Physics - Magnetic Effects of Current: An electric charge $+q$ moves with velocity $\mathbf{v}=3\hat{\mathbf{i}}+4\hat{\mathbf{j}}+\hat{\mathbf{k}}$, in an electromagnetic field given by $\mathbf{E}=3\hat{\mathbf{i}}+\hat{\mathbf{j}}+2\hat{\mathbf{k}}$, $\mathbf{B}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-3\hat{\mathbf{k}}$. The $y$ component of the force experienced by $+q$ is

Navigating Electromagnetic Fields

Imagine a charged particle moving through the vastness of space. Suddenly, it enters a region where both electric and magnetic fields are present. It's no longer just drifting; it's caught in a cosmic dance, pushed and pulled by invisible forces. The total force it experiences in this scenario is beautifully captured by a single, elegant equation known as the Lorentz Force.

Whenever a charge $q$ moves with a velocity $\mathbf{v}$ in a region containing an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$, the net force $\mathbf{F}$ acting on it is given by:

This equation is a combination of two distinct phenomena. The first part, $q\mathbf{E}$, is the electric force. It's straightforward: the force acts parallel to the electric field. The second part, $q(\mathbf{v}\times \mathbf{B})$, is the magnetic force. This one is trickier because it involves a cross product, meaning the force is always perpendicular to both the velocity of the particle and the magnetic field itself.

A Mathematical Right-Hand Rule

To solve our problem, we first need to tackle the magnetic component by calculating the cross product of the velocity and the magnetic field. We are given:

We use the determinant method to find $\mathbf{v}\times \mathbf{B}$. This method ensures we correctly account for the 3D directional nature of the vectors. Setting up the determinant:

Expanding this determinant carefully to avoid any sign errors:

Now that we have the magnetic force component (divided by $q$), we simply add it to the electric field vector $\mathbf{E}$ to find the total force vector (divided by $q$). The electric field is given as:

Adding the two vectors together component by component:

Multiplying the charge $q$ back in, we get the final net force vector:

The question specifically asks for the $y$-component of the force experienced by the charge. In vector notation, the $y$-component corresponds to the coefficient of the $\hat{\mathbf{j}}$ unit vector.

Looking at our final force vector, the coefficient of $\hat{\mathbf{j}}$ is $11q$. Therefore, the $y$-component of the force is $11q$.

Always remember to read the question carefully! If it had asked for the magnitude of the total force, you would have needed to take the square root of the sum of the squares of all three components: $|\mathbf{F}|=q\sqrt{(-10{)}^2+11^2+1^2}=q\sqrt{222}$. Being comfortable with these vector operations is crucial for mastering electromagnetism.