Magnetic Force Direction Explained

This question asks for the direction of the magnetic force experienced by a conductor carrying current in a magnetic field. We can determine this using the fundamental principles of electromagnetism, specifically the Lorentz force law for currents.

Applying the Lorentz Force Law

The magnetic force ($\vec{F}$) acting on a straight conductor of length vector $\vec{L}$ carrying a current $I$ in a uniform magnetic field $\vec{B}$ is given by the formula:

$ \vec{F} = I (\vec{L} \times \vec{B}) $

Defining the Vectors

Let's define the vectors based on the problem statement:

• The conductor is placed along the z-axis, and the current flows in the z-direction. Therefore, the length vector $\vec{L}$ is along the positive z-axis. We can represent this using the unit vector $\hat{k}$: $ \vec{L} = L \hat{k} $ where $L$ is the length of the conductor.
• The uniform magnetic field is directed along the y-axis. We can represent this using the unit vector $\hat{j}$: $ \vec{B} = B \hat{j} $ where $B$ is the magnitude of the magnetic field.

Calculating the Cross Product

Now, we substitute these vectors into the Lorentz force equation:

$ \vec{F} = I ( (L \hat{k}) \times (B \hat{j}) ) $

We can pull the scalar quantities $I$, $L$, and $B$ out of the cross product:

$ \vec{F} = I L B (\hat{k} \times \hat{j}) $

To find the direction, we evaluate the cross product of the unit vectors $\hat{k} \times \hat{j}$. Recall the cyclic order of unit vectors ($\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}$):

• $\hat{i} \times \hat{j} = \hat{k}$
• $\hat{j} \times \hat{k} = \hat{i}$
• $\hat{k} \times \hat{i} = \hat{j}$

From this, we know that reversing the order negates the result:

$ \hat{k} \times \hat{j} = -(\hat{j} \times \hat{k}) = -\hat{i} $

Determining the Force Direction

Substituting the result of the cross product back into the force equation:

$ \vec{F} = I L B (-\hat{i}) $

$ \vec{F} = - I L B \hat{i} $

The vector $-\hat{i}$ represents the direction along the negative x-axis.

Conclusion

Therefore, the magnetic force acting on the conductor is directed along the negative x-axis.