Motional EMF

## **Problem**

A

26.0

c
m

{\displaystyle 26.0 \; cm}

 long metal rod is placed on top of a metal u-shaped track of equal width. The resistances of the metal rod and the metal track are

2.5

Ω

{\displaystyle 2.5 \; \Omega }

 and

25.0

Ω

{\displaystyle 25.0\; \Omega }

 respectively. When a

0.65

T

{\displaystyle 0.65\; T}

 magnetic field directed into the page is turned on, the metal rod moves to the right with a speed of

1.9

m

/

s

{\displaystyle 1.9\; m/s}

.

**Part 1:** Derive the formula to calculate the magnitude of the induced EMF.

**Part 2:** Calculate the induced EMF.

**Part 3:** Determine the current through the u-shaped track.

**Part 4:** What is the external force needed to keep the rod's velocity constant?

## **Solution**

**Part 1**

Start with the general induced EMF formula

E

=
−

d

Φ
B

d
t

{\displaystyle \mathcal{E} = -\frac{d \Phi\_B}{dt}}

where

Φ
B

=
B
×
A

{\displaystyle \Phi\_B = \boldsymbol{B} \times \boldsymbol{A}}

 (cross product of the magnetic field and surface area vector) is the magnetic flux. Given that the magnetic field is perpendicular to the surface area vector,

Φ
B

=

B

A

sin
⁡

90

∘

=
B
A
.

{\displaystyle \begin{align}
&= BA.
\end{align}}

Since the length of the rod and magnetic field strength are constants, the magnetic flux becomes

Φ
B

=
B
l
x
.

{\displaystyle \Phi\_B = Blx.}

Substitute the magnetic flux into the induced EMF formula to get

E

=
−
B
l

d
x

d
t

.

{\displaystyle \mathcal{E} = -Bl\frac{dx}{dt}.
}

Since the velocity is defined as

# v

d
x

d
t

{\displaystyle v = \frac{dx}{dt} }

, we get

E

=
−
B
l
v
.

{\displaystyle \mathcal{E} = -Blv.}

Therefore, the magnitude of the induced emf is

E

=
B
l
v
.

{\displaystyle \mathcal{E} = Blv.}

**Part 2**

Substituting the known values

# B
0.65

T

{\displaystyle B = 0.65\; T}

,

# l
26.0

c
m

{\displaystyle l = 26.0 \; cm}

, and

# v
1.9

m

/

s

{\displaystyle v = 1.9\; m/s}

 into the formula derived in part 1 gives

E

=
(
0.65

T
)
(
0.26

m
)
(
1.9

m

/

s
)

=
0.3211

V
.

{\displaystyle \begin{align}
\mathcal{E} &= (0.65 \;T)(0.26\; m)(1.9 \;m/s) \\[5 pt]
&= 0.3211 \; V.
\end{align}}

**Part 3**

The metal rod and the u-shaped track can be modeled as two resistors connected in series. Thus, the total resistance is

R

=

R

r
o
d

R

c
o
n
d
u
c
t
o
r

=
2.5

Ω
25

Ω

=
27.5

Ω
.

{\displaystyle \begin{align}
R &= R\_{rod} + R\_{conductor} \\[5 pt]
&= 2.5 \; \Omega + 25 \; \Omega \\[5 pt]
& = 27.5 \; \Omega.
\end{align}}

Now use Ohm's law to determine the total current. Remember that for a series circuit, the current is equal across each resistor.

I

=

E

R

=

0.3211

V

27.5

Ω

=
0.01168

A
.

{\displaystyle \begin{align}
I &= \frac{\mathcal{E}}{R}\\[5 pt]
&= \frac{0.3211 \; V}{27.5 \; \Omega} \\[5 pt]
&= 0.01168 \; A.
\end{align}}

Since the magnetic flux is increasing into the page (increasing area), Lenz's law dictates the system will generate a magnetic field directed out of the page to counteract the increasing flux. By the right hanf rule we find the induced current will move in the counterclockwise direction.

**Part 4**

First calculate the power from this configuration.

P

=
I

E

=
(
0.3211

V
)
(
0.01168

A
)

=
0.003749

W

{\displaystyle \begin{align}
P &= I\mathcal{E} \\[5 pt]
&= (0.3211 \; V)(0.01168 \;A) \\[5 pt]
&= 0.003749 \; W
\end{align}}

Since the power is equal to the force time velocity

# P
F
v

{\displaystyle P = Fv}

thus, the force is

# F

P
v

.

{\displaystyle F = \frac{P}{v}.}

Substituting the power and velocity yields

F

=

0.0037

W

0.19

m

/

s

=
0.0195

N
.

{\displaystyle \begin{align}
F &= \frac{0.0037 \; W}{0.19 \; m/s} \\[5 pt]
&= 0.0195 \; N.
\end{align}}