Observe that the given vector field is a gradient field:
Let
, so that



Integrating the first equation with respect to
, we get

Differentiating this with respect to
gives

Now differentiating
with respect to
gives

Putting everything together, we find a scalar potential function whose gradient is
,

It follows that the curl of
is 0 (i.e. the zero vector).
Answer:
as we find the resultant of the magnetic field we get in the xy plane. Just perpendicular to this plane i.e. along z-axis the charge is moving. Hence the force acting on the charge will also be in the xy plane.
Net field = ./(0.078^2 +0.070^2) = 0.105 T
Angle with x axis is arc tan 0.070/0.072 = 44.2 deg
So the force acting on the moving charge is got by Bqv
0.105*5.5*10^-5 * 6*10^3 = 0.03465 N
So the angle inclined by this force with x axis will be 90+44.2 = 134.2
Explanation:
Speed of the TRAX is given as 16 m/s so let say it is given as

now our speed towards the front end is given by 5 m/s so this is the relative speed of us with respect to TRAX
let say this speed is given as

now we need to find the speed with respect to someone standing outside the TRAX
so here we need to find the net speed in ground frame and hence we can use the formula of relative speed




so someone outside the TRAX will see our speed as 21 m/s