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?
At first, we have:
"q=1.24\\times10^{-8}\\,C; \\,\\vec{v}=\\begin{bmatrix}\n1.49\\times 10^1, & -3.85\\times 10^1, & 0,\n\\end{bmatrix}\n\n\\tfrac{m}{s}"
"(a)\\,\\vec{B} = \\begin{bmatrix}\n1.40, & -2.30, & 0,\n\\end{bmatrix} T"
"(b)\\,\\vec{B} = \\begin{bmatrix}\n-1.25, & 2.00, & -3.50,\n\\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}\n\\,\\widehat{i} & \\widehat{j} & \\widehat{k}\\,\\\\ \n14.9 & -38.5 & 0\\\\ \n1.40 &-2.30 & 0\n\\end{pmatrix}\\,N"
"\\vec{F_{B}}=(1.24\\times10^{-8}\\,N)\\begin{pmatrix}\n14.9 & -38.5\\\\ \n1.40 &-2.30\n\\end{pmatrix}\\widehat{k}"
"\\vec{F_{B}}=(1.24*10^{-8}\\,N)((-14.9*2.30)-(-1.40*38.5))\n\\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}\n\\widehat{i} & \\widehat{j} & \\widehat{k}\\\\ \n14.9 & -38.5 & 0\\\\ \n-1.25 &-2.00 & -3.50\n\\end{pmatrix}\\,N" =
"(1.24)(10^{-8})\\,N)\\left [ \\begin{pmatrix}\n14.9 & -38.5 \\\\ \n-1.25 & -2.00\n\\end{pmatrix}\\widehat{k}\\,+\\begin{pmatrix}\n\\widehat{i} & \\widehat{j} \\\\ \n14.9 & -38.5\n\\end{pmatrix}(-3.50) \n\n\\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:
Reference:
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