Question

One coulomb of point charge moving with a uniform velocity \(10\hat{x}\) m/s enters the region \(x \ge 0\) having magnetic flux density \(\vec{B} = (10y\hat{x} + 10x\hat{y} + 10\hat{z})\) T. The magnitude of force on the charge at \(x = 0^+\) is \(\underline{\hspace{2cm}}\) N.

Hint:

Magnetic force depends only on velocity and magnetic field at the instant; electric charge path does not matter.

Answer 1

Correct Answer: 100

Magnetic force on a moving charge:
\[ \vec{F} = q \, \vec{v} \times \vec{B} \] Given:
\[ q = 1\ \text{C}, \vec{v} = 10\hat{x} \] At \(x = 0^+\):
\[ B_x = 10y = 0, B_y = 10x = 0, B_z = 10 \] So:
\[ \vec{B} = 10\hat{z} \] Cross product:
\[ \vec{v} \times \vec{B} = (10\hat{x}) \times (10\hat{z}) \] Using identity \(\hat{x} \times \hat{z} = -\hat{y}\):
\[ \vec{F} = 100 (-\hat{y}) \] Magnitude:
\[ |\vec{F}| = 100\ \text{N} \] \[ \boxed{100\ \text{N}} \]