Chapter 25, Problem 46A

(a)

To determine

To calculate:The vertical component of the magnetic field.

Answer to Problem 46A

  $0.039\text{T}$

Explanation of Solution

Given:

  $B$ is the magnetic field and $\theta$ is the given angle between field and velocity

  $\begin{array}{l}B=0.045\text{T}\\ \theta =60.0°\end{array}$

Formula used:

The vertical component of the magnetic field can be calculated by the formula,

= $B\times \sin \theta$

where,

  $B$ = Magnetic field

  $\theta$ = Angle between field and velocity

Calculation:

Substituting the values in the above formula,

  $\begin{array}{l}=(0.045T)(sin60.0°)\\ =0.039T\end{array}$

Conclusion:

Hence, the vertical component of the magnetic field $=0.039T$ .

(b)

To determine

To calculate:EMF induced in the wire.

Answer to Problem 46A

EMF induced in the wire is $0.23\text{V}$ .

Explanation of Solution

Given:

  $\begin{array}{l}{B}_y=0.039\text{T}\\ L=2.5\text{m}\\ v=2.4\text{m/s}\end{array}$

Formula used:

  $EMF=B_{y}Lv$

where,

  $B_y$ : Vertical component of the magnetic field

  $L$ : Length

  $v$ : Velocity

Calculation:

EMF induced in the wire can be calculated by the formula,

  $EMF=B_{y}Lv$

Substituting the values in the above formula,

  $\begin{array}{l}=\left(0.039\text{T}\right)\left(2.5m\right)\left(2.4m/s\right)\\ =0.23V\end{array}$

Conclusion:

HenceEMF induced in the wire is $=0.23V$ .