Question

F=q(v x B)

q=1.6E-19
v=10^6 m/s in direction -j+k
B=0.0004k T
...so why is v=10^6 ((-j+k)/sqrt(2))? why divide by sqrt(2)?

Answer 1

because V X B= |v||B| sin(theta) where theta is the angle between v and B so , we find that sin(theta) for given case is equal to 1/sqroot(2) bcoz theta=45degree

Answer 2

The problem is an example in my text book. It states that the vector -j+k has length sqrt(2) and v has magnitude 10^6. Therefore v =v=10^6 ((-j+k)/sqrt(2)). but why are we dividing the vector by its length?

Answer 3

9

Answer 4

a vector is always written as vector = magnitude *(unit vector).. but the given v is not of this form... so wat we do is to find a unit vector along the vector -j+k and multiply that by the magnitude..
to find the unit vector along -j+k divide this vector -j+k by its magnitude... here the magnitude is sqroot(2)..
((
you need to understand that we r to simply multiply the magnitude by a unit vetor... and the unit vector is the vector -j+k divided by the magnitude of -j+k ... another way to calculate...
let the velocity be v(vector)= c(-j+k) where c is a constant ... now given that magnitude of the vector v is 10^6.. therefore |v(vector)|=10^6 (implies)=> |c(-j+k)|=10^6.. as c is a contant so this implies c|(-j+k)|=10^6 => c * sqroot(2)=10^6 => c=10^6/sqroot(2); => v(vector)=10^6/sqroot(2) ))