How to find the EMF in a bullet?
At the equator, the earth's magnetic field is approximately horizontal and is directed toward the north and has a value of 8*10^-5 T. Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 cm and a diameter of 0.4 cm and is traveling at 350 m/s.
I have no idea what to do. The equation I used: E= BLV where B is 8*10^-5, L is .01, and V is 350 is wrong. So what else can I do?
Please help, this problem is killing me.
- 1 decade agoFavorite Answer
< Please help, this problem is killing me.>
Nobody said you had to stand in the path of the bullet! :^)
Let a small test charge q rest within the bullet. The force on this charge, which is qE + qv×B, is zero. To counteract the force on the test charge due to the qv×B term, which will be upward for a positive test charge, the electrical field must be uniformly downward. The voltage between the topmost and bottommost part of the bullet is therefore Ed, where d is the diameter of the bullet, the top of the bullet being positive. Because E = vB, the voltage V will be
V = Ed = vBd
= 350 m/s * 8e-5 T * 4e-3 m
= 0.11 millivolts
The length of the bullet doesn't enter into the calculation.
This is also in addition to the speed of the bullet resulting from the earth's rotation, which is roughly comparable in magnitude to the bullet's muzzle velocity.