Question

A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. the horizontal component of the earth's magnetic field at that location is 25 μt toward the north and the vertical component is 50μt downward. what is the emf induced between the ends of the wire?

Answer 1

The emf induced between the ends of the wire is about 50 μV

$\texttt{ }$

Further explanation

Let's recall the induced emf formula for straight wire moving in magnetic field as follows:

$\boxed {\varepsilon = B L v}$

where:

ε = induced emf ( V )

B = magnetic field strength ( T )

L = length of wire ( m )

v = speed of wire ( m/s )

B, L , v are mutually perpendicular

Let us now tackle the problem!

Given:

speed of wire = v = 2.0 m/s

length of wire = L = 50 cm

magnetic field strength = B = 50 μT → vertical component

Asked:

induced emf = ε = ?

Solution:

$\varepsilon = B L v$

$\varepsilon = (50\ \mu T)(50 \ cm)(2.0 \ m/s)$

$\varepsilon = (50\ \mu T)(0.5 \ m)(2.0 \ m/s)$

$\varepsilon = (50\ \mu T)(1.0 \ m^2/s)$

$\varepsilon = 50\ \mu V$

$\texttt{ }$

Conclusion:

The emf induced between the ends of the wire is about 50 μV

$\texttt{ }$

Learn more

• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810

• Compare and contrast a series and parallel circuit : brainly.com/question/539204

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Electromagnetism

Answer 2

When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

$EMF = v.(B x L)$

given that

$B = 25\mu Tj - 50\mu Tk$

$v = 2 m/s j$

$L = 0.50 m (-i)$

now by using the above formula we will have

$EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)$

$EMF = 2(j) .(12.5\mu k + 25\mu j)$

$EMF = 50 \mu Volts$