Question

In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50$\mathit{\mu }$C and initially moving northward atis deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on
the particle.

Short answer

The Magnetic field force is 0.0505N

Step-by-step solution

Important Concepts

The magnetic field force is given by

$$\begin{gathered} {\mathbf{F}}_{\mathbf{B}}\mathbf{=}\mathbf{q}\left(v\times B\right) \\ \mathbf{F}\mathbf{=}\mathbf{qvBsin\theta } \end{gathered}$$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

Application

Using the given data we get

$$\begin{gathered} {\mathrm{F}}_{\mathrm{B}}=\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\mathrm{sin\theta } \\ {\mathrm{F}}_{\mathrm{B}}=\left(8.50\times {10}^{-6}\mathrm{C}\right)\left(1.25\mathrm{T}\right)\left(4.75\times {10}^3\mathrm{m}/\mathrm{s}\right)\sin \left(90\right) \\ {\mathrm{F}}_{\mathrm{B}}=0.0505\mathrm{N} \end{gathered}$$

Using the right hand thumb rule, to v and B, thumb Points to east hence

The magnetic field force is 0.0505N and its direction is towards east.