Finding magnetic field from magnetic force / semielliptical conductor

[awelex]

Homework Statement

1) A charge Q is placed in a uniform magnetic field of magnitude a that proceeds in the -z direction. Its displacement is given by r(t) = (at^2+bt, 2bt, a).
a) Find v(t).
c) Find the magnetic force exerted on the charge at time t=t'.
d) Now the magnetic field at time t' has changed; a magnetic force vector F = (0, -ab, 4bt^2) is now exerted by the field on the charge in addition to the original magnetic force vector. If the charge has the same velocity, and y-component of B = 0, what's the additional magnetic field?

2) A semi-elliptical conductor (semimajor axis a, semiminor axis b) is centered at the origin in the xy-plane. The conductor carries a current I a CCW direction.

Write the integral expression for the magnetic field vector due to the semi-elliptical conductor at the origin.

Homework Equations

The Attempt at a Solution

1)
a) v(t) = r'(t) = (2at + 2b, 2b, 0)

c) not sure about this one:
F = Q * v(t') * B * sin(theta); but theta = 90 degrees because B goes in -z direction and v(t) doesn't have a z component. Therefore,
F = Q * sqrt( (2at+2b)^2 + 4b^2 ) * a

d) This is the one I have no idea how to solve. I don't even know how to start. Do I have to use the law of Biot and Savart to find the magnetic field? If so, how? I assume another possibility would be to find the "inverse cross product" (if such a thing exists) of the given force vector - but I was told by our professor we don't have to do that. Any hints?


2) We only have to express the integral, not solve it. This is how far I got:

dB = mu0 / (4*Pi) * (I * dl) / r^2, because dl is perpendicular to the "radius" of the ellipse at any point. But the radius itself varies. How do I express that? I looked up the the equations of an ellipse in polar coordinates, so I assume I could use that. But I'd rather understand what exactly I'm doing than simply plugging in equations from the internet.

Thanks.

 

[saubhik]

1.d)biot and savart is for current distributions; why use that?

Go on with the lorentz force law. Vectorially add the magnetic forces and equate to (qv)X(B_net). You'll get 3 equations each along x,y and z directions. Solve them and get your field B.

regarding 2, what is the centre of a semi-elliptical conductor?

 

[awelex]

Thanks for your answer, saubhik. For 1d) I assumed that's how I have to solve it, but I was specifically told not to - there must be some property that simplifies the problem.

The center of the semi-elliptical conductor is the the point midway between the two foci.