How Is Induced EMF in a Moving Car Antenna Calculated?

[kmj9k]

A car with a vertical radio antenna 75 cm long drives due east at 29 m/s. The Earth's magnetic field at this location has a magnitude of 5.9* 10^-5 T and points northward, 72° below the horizontal.
(b) Find the induced emf between the ends of the antenna.

Relevant Equations:
E = BVL



I just tried to plug into E=BVL, using B as (5.9 * 10^-5), V as 29 m/s, and L as 0.75m. I ended up with 1.28 mV, which I know is wrong. But, this problem seems so simplistic, I don't understand what I'm doing wrong.

 

[apelling]

A simple correction is needed for the 72 degree angle, but it takes a bit of explaining why it is needed:-

Induced Emf is equal to flux linkage change per second from Faraday.

VL gives you a rectangle that is equal to the area of space swept by the Aerial in 1 second.

Through this rectangle magnetic flux is flowing. This flux is cut by the aerial as it sweeps by. Emf = flux cut per second.

Flux = B*Area (this will be the flux cut second so is also the Emf we are after) This is where BVL comes from.

But for this equation to work the flux must be at right angles to the area. If not we need to find the component of the flux density which is at right angles to the area.

This requires a simple bit of trigonometry and this is where the bit about the Earth's field being angled down comes in.

Try drawing this rectangle and putting on the flux vector and seeing what correction is needed for the 72 degrees angle.
(If the angle where 0 degrees all the flux is cut no correction is needed, if it where 90 degrees no flux is cut)

Hopefully that made sense and you can now correct your answer.

 

[kmj9k]

I got it! I used B*Area (or BVL) times the cos of 108 and got 0.397 mV. Thank you so much for your help!