Answer:
D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Explanation:
The statement is not correctly written, the correct form is now described:
A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?
From classic theory on Magnetism, we remember that the magnetic force exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:
[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)
Where:
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in teslas.
If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]
[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Which corresponds to option D.
