Chapter 25, Problem 45A

(a)

To determine

To Calculate:The potential difference induced in the given conductor.

Answer to Problem 45A

  $0.13\text{V}$

Explanation of Solution

Given:

Length of the wire, $L=0.50\text{m}$

Magnetic Field, $B=7.0\times {10}^{-2}\text{T}$

Speed of the wire, $v=3.6\text{m/s}$

Resistance, $R=11\Omega$

  $\theta =90°$

Formula used:

The EMF is:

  $EMF=BLv\left(\sin \theta \right)$

Where,

  $EMF$ : Electromotive Force

  $B$ : Magnetic Field

  $L$ : Length

  $v$ : Velocity

  $R$ : Resistance of the circuit

Calculation:

Substitute all the values in the given formula,

  $EMF=\left(\text{7}{\text{.0×10}}^{\text{-2}}\text{T}\right)\text{×}\left(\text{0}\text{.50}\text{m}\right)\text{×}\left(\text{3}\text{.6}\text{m/s}\right)\text{×}\left(\sin 90°\right)$

  $EMF=0.13\text{V}$

Conclusion:

Hence, the potential difference induced in the given conductor is $0.13\text{V}\text{.}$

(b)

To determine

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

Answer to Problem 45A

  $0.011\text{A}$

Explanation of Solution

Given:

The potential difference induced in the given conductor is $0.13\text{V}\text{.}$

Resistance of circuit is $11\Omega$ .

Formula used:

  $I=\frac{EMF}{R}$

Where,

  $EMF$ : Electromotive Force

  $R$ : Resistance of the circuit

  $I$ : Current

Calculation:

The formula to calculate the current is given by,

  $I=\frac{EMF}{R}$

Substituting all the values in the given formula,

  $I=\frac{\text{0}\text{.13}\text{V}}{\text{11}\text{Ω}}$

  $I=0.011\text{A}$

Conclusion:

Hence, the current in the circuit is $0.011\text{A}$ .

(c)

To determine

To Explain:The polarity of point A relative to point B.

Answer to Problem 45A

Point A is negative with relative to point B.

Explanation of Solution

Introduction:

The current flows from positive terminal to negative terminal. The polarity can be determined by finding the direction of the current in the conductor.

As per Fleming’s left hand rule, the thumb points towards right, middle finger points towards magnetic field which upwards, the index finger points from top to bottom. So, point A is positive relative to point B.

Conclusion:

Hence,point A is negative in relation to the point B.