Chapter 25, Problem 39A

To determine

To Calculate: The speed for length of wire.

Answer to Problem 39A

  $20\text{m/s}$

Explanation of Solution

Given:

EMF, $10\text{V}$

Magnetic Field, $B=2.5\text{T}$

Length of wire, $l=0.2\text{0}\text{m}$

  $\theta =90°$

Formula Used:

EMF $=$ $Blv\left(\sin \theta \right)$

Where,

  $B$ : Magnetic Field

  $l$ : Length

  $v$ : Velocity

Calculation:

Re-arrange the formula to calculate the velocity,

  $v=\frac{EMF}{BL\left(\sin \theta \right)}$

Now,substitute all the values in the given formula,

  $\begin{array}{l}v=\frac{\text{10}\text{V}}{\left(\text{2}\text{.5}\text{T}\right)\times \left(\text{0}\text{.20}\text{m}\right)\times \left(\sin 90°\right)}\\ v=20\text{m/s}\end{array}$

Conclusion:

Hence the speed at which $0.20\text{m}$ length of wire perpendicularly across $2.\text{5}\text{T}$ magnetic field is $20\text{m/s}$ .