Question

A particle with a charge of -1.24 x 10 C is moving with instantaneous velocity = (1.49 x 101)+(-3.85 x 101. What is the force exerted on this particle by a magnetic field (a) = (1.40 7) + (-2.30 7) and (b) B = (-1.25T) + (2.007)j + (-3.50 T)k?

Answer 1

At first, we have:

$q=1.24\times10^{-8}\,C; \,\vec{v}=\begin{bmatrix} 1.49\times 10^1, & -3.85\times 10^1, & 0, \end{bmatrix} \tfrac{m}{s}$

$(a)\,\vec{B} = \begin{bmatrix} 1.40, & -2.30, & 0, \end{bmatrix} T$

$(b)\,\vec{B} = \begin{bmatrix} -1.25, & 2.00, & -3.50, \end{bmatrix}T$

$\vec{F} = q\cdot \vec{v}\times \vec{B}$

Using the definition of the Lorentz force will allow us to calculate the force exerted by each magnetic field, thus we find that:

(a) $\vec{F_{B}}=(1.24\times10^{-8}) \begin{pmatrix} \,\widehat{i} & \widehat{j} & \widehat{k}\,\\ 14.9 & -38.5 & 0\\ 1.40 &-2.30 & 0 \end{pmatrix}\,N$

$\vec{F_{B}}=(1.24\times10^{-8}\,N)\begin{pmatrix} 14.9 & -38.5\\ 1.40 &-2.30 \end{pmatrix}\widehat{k}$

$\vec{F_{B}}=(1.24*10^{-8}\,N)((-14.9*2.30)-(-1.40*38.5)) \widehat{k}$

$(a) \vec{F_{B}}=(2.43\times10^{-7}\,N)\,\widehat{k}$

Then, for b:

(b) $\vec{F_{B}}=(1.24\times10^{-8}) \begin{pmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 14.9 & -38.5 & 0\\ -1.25 &-2.00 & -3.50 \end{pmatrix}\,N$ =

$(1.24)(10^{-8})\,N)\left [ \begin{pmatrix} 14.9 & -38.5 \\ -1.25 & -2.00 \end{pmatrix}\widehat{k}\,+\begin{pmatrix} \widehat{i} & \widehat{j} \\ 14.9 & -38.5 \end{pmatrix}(-3.50) \right ]$

=$(1.24*10^{-8})[(-29.8-48.125)\widehat{k}\,+(-38.5\widehat{i}-14.9\widehat{j})(-3.50)]$

= $(134.75\widehat{i}+52.15\widehat{j}-77.925\widehat{k})(1.24*10^{-8})\,N$

(b) $\vec{F_{B}}=(16.71\widehat{i}+6.47\widehat{j}-9.66\widehat{k}\,) *10^{-7}\,N$

In conclusion, the forces exerted on the charged particle due to the magnetic field are:

• $(a)\,\vec{F_{B}}=(2.43*10^{-7}\,N)\,\widehat{k}$
• $(b)\,\vec{F_{B}}=(16.71\widehat{i}+6.47\widehat{j}-9.66\widehat{k}\,) *10^{-7}\,N$

Reference:

• Young, H. D., & Freedman, R. A. (2014). Sears and Zemansky's University Physics (with Modern Physics Technology Update). Edinburgh: Pearson.