Shooting a bullet into an MRI

Question

What happens, if one shoots a bullet into an MRI system? Is there any chance to fully stop the bullet?

Boundary condifitions:

Consider an MRI machine with a reasonalby strong magnetic field (~3T) and tunable gradient coils. Furthermore, suppose we have a standard gun and bullet to shoot into the field of the machine.

Can we find a trajectory, that stops the bullet from leaving the MRI system?

For example, stopping it, reversing it, leading it on a circular path inside the machine etc. etc.?

Update: The stationary field inside the MRI machine seems to be a bad candidate, as it follows the centerline in a parallel fashion. So, is there any chance of eddy currents inside the bullet that help slowing it down? I found the x-y-gradient fields in a typical MRI system, the peak gradient is about 35-45 mT/m.The slew rate, which is defined as peak gradient over rise time can reach up to 200 T/m/s.

Answer 1

New Version (as I had my doubts on the orders of magnetude...and it turned out that I made a mistake)

To simplify things I assume:

1. velocity $v=1000\; \mathrm m /\mathrm s$
2. perfectly conducting bullet
3. density of $\rho=10000\; \mathrm{kg}/\mathrm{m}^3$
4. A $B_\mathrm{max}=5\;\mathrm T$ and constant gradient of $B'=5\;\mathrm{T}/\mathrm{m}$

If the bullet is a considered a single winding solenoid field inside is

$$B=\mu_0 \lambda(x),$$ where $\lambda$ is the surface charge density. The dipole moment is $$P=\mu_0 A I=\mu_0 V \lambda,$$ with current $I=l\lambda$,length $l$, area $A$, and Volume $V$. In this geometry the force on the dipole is $$F(x)=\frac{P}{\mu_0} \frac{\mathrm d B}{\mathrm d x}=\frac{\mu_0 V \lambda}{\mu_0} B'=\frac{V B(x)}{\mu_0} B'$$ and as $B(x)=xB'$,i.e. constant gradient $$F(x)=V\frac{(B')^2}{\mu_0}x$$ the integrated energy then is $$E(x)=V\frac{(B')^2}{\mu_0}\frac{x^2}{2}$$ and we should have $$V\frac{(B')^2}{\mu_0}\frac{x^2}{2}=\frac 1 2 m v^2 = \frac{\rho V}{2}v^2$$ hence $$\frac{(B')^2}{\mu_0}x^2=\rho v^2$$ i.e. $$x=\frac{v}{B'}\sqrt{\mu_0 \rho},$$ which in our case is $22\;\mathrm m$ and not one.

Again looking at energy we could try to figure out the speed reduction $$v=\sqrt{v_0^2-\frac{(B')^2}{\rho \mu_0}x^2}$$ which is insignificant. Moreover I guess that there are no reasonable $B'$ and $\rho$ to make it work. But you could probably stop a $150\;\mathrm m/\mathrm s$ aluminum arrow shot by an archer.

I hope that I have my math correct now.