Question

A 1.70 m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 14.4 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. magnitude mV

Answer 1

Answer:

The Induced emf between the ends of the wire is 0.0009792V.

Explanation:

Since

The Induced EMF is defined as

ε = B L v

where B Earth's  magnetic field in the reggion, L is the length of wire and v is the speed .

so

ε = ($40*10^-6$)(1.70)(14.44)

by multiplying these values we get

ε =  0.0009792V

so that is  induced emf

Answer 2

Answer:

The induced emf is 0.9792 mV.

The direction of the field indicates the position of point charge, hence the downward end is positive.

Explanation:

Given;

length of wire, L = 1.70 m

speed of the conductor, v = 14.4 m/s

strength of magnetic field, B = 40.0 μT

Induced emf can be calculated by applying Maxwell equation of electromagnetic induction,

Induced emf = BLV

where;

B is the strength of magnetic field

L is the length of the conductor

V is the speed of the conductor

Substitute the given values and solve for Induced emf

Induced emf = BLV

                     = 40 x 10⁻⁶ x 1.7 x 14.4

                     = 9.792 x 10⁻⁴ V

The induced emf is 0.9792 mV.

The direction of the field indicates the position of point charge, hence the downward end is positive.