Question

A 1.95 m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 15.5 m/s. The vertical component of Earth's magnetic field in this region is 40.0 μT directed downward. Calculate the induced emf between the ends of the wire and determine which end is positive.

Answer 1

`\epsilon=1.2* 10^(-3)\ V`

Step-by-step explanation:

Given that,

Length of a wire, l = 1.95 m

Speed of the wire, v = 15.5 m/s

The vertical component of Earth's magnetic field in this region is 40.0 μT directed downward.

We need to find the induced emf between the ends of the wire and determine which end is positive. The induced emf for a wire is given by :

`\epsilon=Blv\\\\\epsilon = 40* 10^(-6)* 1.95* 15.5\\\\\epsilon=1.2* 10^(-3)\ V`

So, the induced emf between the ends of the wire is
`1.2* 10^(-3)\ V`.