A potential difference of 38102 V is developed across a 10cm

A potential difference of 3.8×10?2 V is developed across a 10-cm -long wire as it moves through a magnetic field at 5.0 m/s. The magnetic field is perpendicular to the axis of the wire and to the direction of motion.

a.) What is the strength of the magnetic field? (Answer in T units)

Solution

The motional EMF is given by E = (v x B)*L

Since the axis of the wire is perpendicular to the magnetic field, this is E = v*B*L

=> B = E/(v*L)

= (3.8*10^-2)V/(5m/s)*(0.1m)

= 7.6*10^-2T (Answer)