Question

A positive point charge with velocity \(\vec v=5\hat x\) enters a region having electric field \(\vec E=4\hat y\) and magnetic field \(\vec B=-6\hat z\). Which one of the following statements is correct for the force on the charge as it enters the region?

• A. The force will be along the magnetic field but perpendicular to the electric field
• B. The force will be along the electric field but perpendicular to the magnetic field
• C. The force will be perpendicular to both electric and magnetic field
• D. The magnetic field does not exert any force on the charge

Hint:

Apply F = q(E + v cross B), compute v cross B first, then compare the resulting direction with E and with B.

Answer 1

The Correct Option is B

Instead of a determinant, we can use the cyclic unit-vector cross product rules directly: $\hat x\times\hat y=\hat z$, $\hat y\times\hat z=\hat x$, $\hat z\times\hat x=\hat y$, and reversing the order of any pair flips the sign, so $\hat x\times\hat z=-\hat y$.

Apply this to $\vec v\times\vec B$:

\[ \vec v\times\vec B=(5\hat x)\times(-6\hat z)=-30(\hat x\times\hat z)=-30(-\hat y)=30\hat y \]

Add the electric field term to get the total force per unit charge:

\[ \vec E+\vec v\times\vec B=4\hat y+30\hat y=34\hat y \]

So the total force is $\vec F=34q\hat y$.

Comparing directions: the result $34q\hat y$ has a component only along $\hat y$, exactly the direction of $\vec E=4\hat y$, and $\hat y$ is orthogonal to $\hat z$, the direction of $\vec B$. So the force is along $\vec E$ and perpendicular to $\vec B$.

\[ \boxed{\text{Force is along}\ \vec E,\ \text{perpendicular to}\ \vec B} \]