# Will my data drives be safe in an emp event? [closed]

If there was an emp event (maybe because the rotation of magnetism of the earth or an atomic bomb) would the data on my hard drives, ssd, dvd, mini dvi tapes and cassette tapes be safe?

• If the emp event is due to a nuke, your very lowest priority problem will be your data integrity. – David White Oct 22 '15 at 1:19

There is no way a switch in magnetism of the Earth is going to cause an EMP event nearly strong enough to pose a threat to data storage. The swiftest switch recorded in ferromagnetic ores is the Laschamp Event (see my answer here) about 41000 years ago, and this took hundreds of years to complete. Given the Earth's magnetic field is of the order $B_\oplus=5\times 10^{-5}{\rm tesla}$, the voltage induced in even a one meter squared conducting loop, calculated by Faraday's law, when this value flips smoothly over one hundred years (say, $10^9$ seconds) is utterly negligible.
Nuclear weapon bursts are a different matter. I am not expert in the phenomenon, but a quick search gives induced voltages of the order of $10^5{\rm V\,m^{-1}}$ (see here). It would be fairly easy to engineer a magnetic hard drive to withstand this level, although, by the same token, careless design might damage the control circuitry and induce arcing between the drive and its case. The magnetic field itself from such a pulse $B=E/c$ will probably not be strong enough to corrupt magnetic data; a plane wave with $10^5{\rm V\,m^{-1}}$ electric field has a magnetic induction of $10^5{\rm V\,m^{-1}}/c\approx 3\times 10^{-4}{\rm T}$ . From here I glean estimates of hundreds of oersteds (magnetic fields) or hundreds to thousands of microtesla (magnetic induction) as conservative safe upper limits on hard drives.
SSD devices are another matter: a $10^5{\rm V\,m^{-1}}$ can easily induce unsafe voltages in a system not engineered to withstand severe electromagnetic interference. I am not expert in ESD safety, so I suggest asking about solid state memory on Electrical Engineering Stack Exchange.
• @ChrisWhite The flux through the loop is $5\times10^{-5}{\rm weber}$, its mean rate of change therefore $5\times10^{-5}{\rm weber}/10^9{\rm s} = 5\times10^{-14}{\rm V}$ as the nett EMF around the loop. – WetSavannaAnimal Oct 22 '15 at 4:17