A helicopter hovers above the north magnetic field pole in a magnetic filed of magnitude 0.5G perpendicular to the ground. The helicopter rotors are 10m long, are made of aluminum, and rotate about the hub with a rotational speed of 10 000 rpm. What is the potential difference form the hub of the to to the end?

I don't know what to do for this problem , what I first did was to transform rpm to m/s

2*pi*10m*10 000 rpm*(1min/60s)

and got 10742 m/s

I don't know where to go from there.

To find the potential difference from the hub of the helicopter rotor to the end, we can start by calculating the induced voltage in the aluminum rotor due to its rotation in the Earth's magnetic field.

The induced voltage (V) can be calculated using the formula:

V = B * L * v

where:
B is the magnetic field strength (0.5 G in this case),
L is the length of the rotor (10 m in this case), and
v is the velocity of the rotor relative to the magnetic field.

You correctly converted the rotational speed from rpm to m/s and obtained 10742 m/s. This is the velocity of the rotor relative to the magnetic field.

Now, we can substitute the values into the formula:

V = 0.5 G * 10 m * 10742 m/s

First, let's convert the magnetic field strength from Gauss to Tesla (1 T = 10,000 G):

V = 0.5 G * 10 m * 10742 m/s * (1 T / 10,000 G)

V = 0.00005 T * 10 m * 10742 m/s

V = 5.371 T⋅m/s

Now, the potential difference (V) is equal to the induced voltage. So, the potential difference from the hub to the end of the rotor is approximately 5.371 volts.

Please note that this calculation assumes a uniform magnetic field and neglects any resistance in the rotor.