Chapter 25, Problem 38A

(a)

To determine

To Calculate:The EMF induced in the wire.

Answer to Problem 38A

  $3.6\text{V}$

Explanation of Solution

Given:

Length of the wire, $l=0.75\text{m}$

Magnetic field, $B=0.30\text{T}$

Speed of the wire, $v=16\text{m/s}$

Angle between motion of wire and magnetic field, $\theta =90°$

Formula used:

Induced EMF is:

  $e=Blv\left(\sin \theta \right)$

Where,

  $B$ : Magnetic Field

  $l$ : Length

  $v$ : Velocity

Calculation:

  $e=Blv\left(\sin \theta \right)$

Substituting all the values in the given formula

  $\begin{array}{l}e=\left(\text{0}\text{.30}\text{T}\right)\text{×}\left(\text{0}\text{.75}\text{m}\right)\text{×}\left(\text{16}\text{m/s}\right)\text{×}\left(\sin \theta \right)\\ e=3.6\text{}\text{V}\end{array}$

Conclusion:

Hence, the induced emf in the wire is $3.6\text{V}\text{.}$

(b)

To determine

To Calculate:The current in the circuit if the total resistance of circuit is $11\Omega$

Answer to Problem 38A

  $0.33\text{A}$

Explanation of Solution

Given:

Induced EMF, $e=3.6\text{V}$

Resistance of circuit, $R=11\Omega$ .

Formula used:

Current in the circuit:

  $I=\frac{e}{R}$

Where,

  $e$ : Electromotive Force

  $R$ : Resistance of the circuit

  $I$ : Current

Calculation:

Substituting all the values in the given formula,

  $I=\frac{\text{3}\text{.6}\text{V}}{11\Omega }$

  $I=0.3\text{3}\text{A}$

Conclusion:

Hence, the current in the circuit is $0.33\text{A}$ .