Question

An automobile has a vertical radio antenna 1.40 m long. The automobile travels at 56.0 km/h on a horizontal road where Earth's magnetic field is 50.0 µT, directed toward the north and downward at an angle of 65.0° below the horizontal.

Calculate the magnitude of this induced emf.

...........MV

Answer 1

The induced emf of a conductor L moving with velocity V in a magnetic field B is given by the formula;

E = (V cross B) dot ( L )

Where V , B & L are vectors.

In your case, I would imagine a plane tangent to the Earth at your location. Let x be the horizontal East/West direction. Let y be the North/South direction and Let z be perpendicular to the plane or the ground.

The unit vectors in these directions are (i, j, k )

Then from the given information your vectors are;

V = (Vx)i and L = (Lz)k and B = (By)j - (Bz)k (minus for below horizontal)

Now (V cross B) = (Vx)(By)(i cross j) - (Vx)(Bz)(i cross k)

= (Vx)(By)k + (Vx)(Bz)j

Now Dot this into (L)k and only the first term is non zero;

(V cross B) dot (L) = (Vx)(By)(Lz)

Where; Vx = 56 km/h = 15.55 m/s

Lz = 1.40 m

By = (B)Sin(65) = (50.0 x 10^-6)(.91) = 45.31 x 10^-6 T

E = (15.55)(1.40)(45.31 x 10^-6) = .000986 volts

I hope your familiar with vector products between unit vectors.