# A 2.00 m length of wire is held in an east-west direction and moves horizontally to the north...

## Question:

A 2.00 m length of wire is held in an east-west direction and moves horizontally to the north with a speed of 15.0 m/s. The vertical component of Earth's magnetic field in this region is 40.0 {eq}\mu {/eq}T directed downward. Calculate the induced emf between the ends of the wire and determine which end is positive.

## Magnetic field:

In physic, the magnetic field is the vector field that produces around the magnetic materials. The magnetic field is co-related with the electric field that is Magnetic field always acts perpendicular to the electric field. The magnetic field lines are regularly moved from the positive terminal to the negative terminal.

Given data

• The length of wire is: {eq}L = 2\;{\rm{m}} {/eq}.
• The speed of the wire is: {eq}v = 15\;{\rm{m/s}} {/eq}.
• The magnetic field is: {eq}B = 40\;{\rm{\mu T}} = 40 \times {10^{ - 6}}\;{\rm{T}} {/eq}.

The induced emf between ends of wire is calculated as,

{eq}E = BLv {/eq}

Substitute the known values.

{eq}\begin{align*} E &= 40 \times {10^{ - 6}}\;{\rm{T}} \times 2\;{\rm{m}} \times 15\;{\rm{m/s}}\\ &= 0.0012\;{\rm{V}} \end{align*} {/eq}

Suppose a positive charge is driving north in a downward magnetic field. By applying the right-hand rule, the direction of force is west. So the west end of the wire will have a net positive charge.

Thus, the induced emf is 0.0012 V. and west end is the positive end.