   # E.m.f. induced in a straight conductor

When a straight conductor is moved through a magnetic field an e.m.f. is induced between its ends. This movement must be in such a direction that the conductor cuts through the lines of magnetic flux, and will be a maximum when it moves at right angles to the field (Figure 1(a)). Let the length of the conductor be L and the flux density of the field be B.
If the conductor moves with velocity v at right angles to the field then the flux cut per second will be BvL (since the conductor will sweep out an area vL every second).

But the rate of cutting flux is equal to the e.m.f. induced in the conductor. Therefore

Induced e.m.f (E) = BLv

If the conductor cuts through the flux at an angle q ((b) in Figure 1), where q is the angle between the magnetic field and the direction of motion, the equation becomes

Induced e.m.f (E) = BLv sinq

You can see that the maximum e.m.f is generated when the conductor moves at right angles to the field.
(q = 90o and so sinq = sin90 = 1).

Example problems
1. Calculate the e.m.f. generated between the wing tips of an aircraft that is flying horizontally at 200 ms-1 in a region where the vertical component of the Earth's magnetic field is 4.0x10-5 T , if the aircraft has a wingspan of 25 m.

E = 4x10-5 x 25 x 200 = 0.2 V

2. Calculate the e.m.f. generated between the fixed and the free ends of a helicopter blade 9.45 m long that is rotating at 3.5 revs per second. The vertical component of the Earth's field has a flux density of 4.0x10-5 T.

Area swept out per second = pR2 x 3.5 = px 9.452 x 3.5
E.m.f. = BAN = 4.0x10-5 x p x 89.3 x 3.5 = 3.93x10-2 V = 39.3 mV.