Chapter 25.1, Problem 3PP

(a)

To determine

To calculate: Magnetic field of the straight wire.

Answer to Problem 3PP

Magnetic field of the straight wire is $0.10\text{T}$ .

Explanation of Solution

Given:

For a straight wire,

Length, $l=30.0\text{m}$

Velocity, $\nu =2.0\text{m/s}$

Induced $EMF$ , $EMF=6.0\text{V}$

Formula used:

EMF is calculated by,

  $EMF=Blv\left(\sin \theta \right)$

Where,

  $l=$ Length of the wire

  $B=$ Magnetic field

  $v=$ Velocity

Calculation:

In the first step by using the formula of $EMF$ ,

  $EMF=Blv\left(\sin \theta \right)$

Substituting the given values in the above formula,

  $\begin{array}{l}EMF=Blv\left(\sin \theta \right)\\ 0.6\text{V}=B\left(30.0\text{m}\right)\left(2.0\text{m/s}\right)\left(1\right)\\ B=0.10\text{T}\end{array}$

Conclusion:

Hence, magnetic field of the straight wire is $0.10\text{T}$ .

(b)

To determine

To calculate: The current in the wire.

Answer to Problem 3PP

The current of a straight wire is $1.2\text{A}$

Explanation of Solution

Given:

Induced $EMF,EMF=6.0\text{V}$

Resistance, $R=5.0\Omega$

Formula used:

  $I=\frac{EMF}{R}$

Where,

  $I=$ Current

  $R=$ Resistance

  $EMF=$ Electromotive force

Calculation:

The formula for current used is,

  $I=\frac{EMF}{R}$

Substituting the values in above formula,

  $\begin{array}{l}I=\frac{EMF}{R}\\ I=\frac{6.0\text{V}}{5.0\Omega }=1.2\text{A}\end{array}$

Conclusion:

Hence, the current of a straight wire is $1.2\text{A}$