Chapter 25, Problem 87A

(a)

To determine

To calculate:

  $EMF$ induced in the wire.

Answer to Problem 87A

  $6.4\text{V}$

Explanation of Solution

Given:

Velocity of wire, $v=8.\text{0}\text{m/s}$

Strength of the magnetic field, $B=\text{2}\text{.0}\text{T}$

Length of the wire, $L=0.4\text{0}\text{m}$

Formula used:

  $EMF=BLv\left(\sin \theta \right)$

Where,

  $EMF:$ Electromotive force

  $B:$ Magnetic field

  $L:$ Length

  $v:$ Velocity

Calculation:

As the wire cuts perpendicularly across the magnetic field, $\sin \theta$

  $=1$

  $EMF$ induced in the wire can be calculated as follows:

  $EMF=BLv\left(\sin \theta \right)$

Substitute the values in above formula

  $\begin{array}{c}EMF=\left(\text{2}\text{.0}\text{T}\right)\left(\text{0}\text{.40}\text{m}\right)\left(\text{8}\text{.0}\text{m/s}\right)\left(\text{1}\right)\\ =\text{6}\text{.4V}\end{array}$

Conclusion:

Hence $EMF$ induced in the wire is $6.4\text{V}$ .

(b)

To determine

To Calculate:Size of current in the wire.

Answer to Problem 87A

  $1.0\text{A}$

Explanation of Solution

Given:

EMF induced in the wire, $EMF=6.4\text{V}$

Resistance of the wire, $R=6.4\Omega$

Formula used:

  $EMF=IR$

Where,

  $EMF:$ Electromotive force

  $I:$ Current

  $R:$ Resistance

Calculation:

  $EMF=IR$

Rearranging the above formula to get $I$ :

  $I=\frac{EMF}{R}$

Substituting the values in above formula:

  $\begin{array}{c}I=\frac{\text{6}\text{.4}\text{V}}{6.4\Omega }\\ =1.0\text{A}\end{array}$

Conclusion:

Hence,current in the wire is $1.\text{0}\text{A}$ .