Chapter 21, Problem 17P

To determine

(a)

The induced emf.

Answer to Problem 17P

Solution:

Induced emf is $\text{0}\text{.168 V}$

Explanation of Solution

Given info:

Magnetic field(B)= $\text{0}\text{.35 T}$

Length (l)= $\text{30}\text{.0 cm}$

= $\text{0}\text{.300 m}$

Speed v = $\text{1}\text{.6 }\text{m}/\text{s}$

Resistance(R)= $\text{2}\text{.5 Ω+21 Ω}$

= $\text{23}\text{.5 Ω}$

Formula to be used:

By Faraday’s law, an induced emf E whose magnitude is given by,

$E=Blv$ [Because velocity is perpendicular to the magnetic Field]

Calculation:

Applying values,

$=\text{ 0}\text{.35}\times 0.300\times 1.6$

$=\text{0}\text{.168 V}$

To determine

(b)

The current in the U-shaped conductor.

Answer to Problem 17P

Solution:

Current in the conductor ${\text{7×10}}^{\text{-3}}\text{ A}$

Explanation of Solution

Given info:

Emf, $E$$=\text{0}\text{.168 V}$

$\text{R=23}\text{.5 Ω}$

Formula to be used:

By ohm’s law, the induced current is given by,

$I=\frac{E}{R}$

Calculation:

Applying values,

$\begin{array}{l}I=\frac{0.168}{23.5}\\ =\text{0}\text{.007 A}\\ \left(or\right)\\ I={\text{7×10}}^{\text{-3}}\text{ A}\end{array}$

To determine

(c)

The external force.

Answer to Problem 17P

Solution:

External force is $\text{7}{\text{.35×10}}^{\text{-4}}\text{ N}$

Explanation of Solution

Given info:

Current, $I={\text{7×10}}^{\text{-3}}\text{ A}$

Magnetic field (B)= $\text{0}\text{.35 T}$

Length (l)= $\text{30}\text{.0 cm}$

= $\text{0}\text{.300 m}$

To keep the rod’s velocity constant, there should be an equal external force to the right because the current is in the upward magnetic field and the magnetic force to the left.

Formula to be used:

Hence to calculate the external force F (for a constant $\overrightarrow{B}$)

$F_{ext}=Ilb$

Calculation:

Applying values,

$\begin{array}{l}{F}_{ext}={\text{7×10}}^{\text{-3}}\times 0.300\times 0.35\\ F_{ext}=\text{7}{\text{.35×10}}^{\text{-4}}\text{ A}\end{array}$