Question

An proton moves with a velocity of bold v with bold rightwards harpoon with barb upwards on top equals open parentheses 5.00 space bold i with bold hat on top minus 6.00 space bold italic j with hat on top plus 1.00 space k with hat on top close parentheses space space straight m divided by straight s in a region in which the magnetic field is bold italic B with bold rightwards harpoon with barb upwards on top equals open parentheses 1.00 space bold i with bold hat on top plus 2.00 space bold italic j with hat on top minus 1.00 space k with hat on top close parentheses space space straight T. The electric charge carried by a proton is 1.60 cross times 10 to the power of negative 19 end exponent space straight C. What is the magnitude of the magnetic force this proton experiences

Answer 1

Answer:

2.64*10^-18 N

Explanation:

You have the following vectors:

$\vec{v}=5.00\hat{i}-6.00\hat{j}+1.00\hat{k}\\\\\vec{B}=1.00\hat{i}+2.00\hat{j}-1.00\hat{k}$

The magnitude of the magnetic force is given by the following equation:

$|\vec{F_B}|=q|\vec{v}\ X\ \vec{B}|$

Thus, you calculate the cross product between v and B (for example with the determinant method) and you obtain:

$\vec{v}\ X\ \vec{B}=4\hat{i}-\hat{j}+16\hat{k}$

Finally the magnitude of the force will be:

$|\vec{F_B}|=(1.6*10^{-19}C)\sqrt{(4)^2+(-1)^2+(16)^2}N=2.64*10^{-18}N$

then, the magnetic force is 2.64*10^-18 N