Chapter 20, Problem 5P

To determine

Part (a) To Determine:

Type of magnetic pole at top face when current carrying wire is observed to jump towards the observer.

Answer to Problem 5P

Solution:

- South Pole

Explanation of Solution

Using the concept of Fleming’s Left-Hand Rule, the wire would jump towards the observer, if the field is going from bottom to top.

To determine

Part (b) To Determine:

The value of current in the wire placed in magnetic field

Answer to Problem 5P

Solution:

- Current I=3.86 A

Explanation of Solution

Given:

-$F_{\max }=8.50\times {10}^{-2}\text{N}$; $l=10\text{cm}$; $B=.220\text{T}$

Formula Used:

$F=Bil\mathrm{Sin}\theta$

Calculation:

Since Force is maximum so

$\theta ={90}^o$

$i=\frac{F_{\max }}{Bl}$

$i=\frac{8.50\times {10}^{-2}}{(.220)(10\times {10}^{-2})}$

$i=3.86\text{A}$

To determine

Part (c) To Determine:

New force on a current carrying conductor when it is tapped

Answer to Problem 5P

Solution - Force = 0.0833 N

Explanation of Solution

Given:

-$F_{\max }=8.50\times {10}^{-2}\text{N}$; $\theta ={10}^o$

Formula Used:

$F=Bil\mathrm{Sin}\theta$

Calculation:

If wire is tipped and makes 10⁰ with the horizontal, then new force will be

$F^{\text{'}}=F_{\max }\mathrm{Cos}{10}^0$

$F^{\text{'}}=(8.50\times {10}^{-2})(0.98)$

$F^{\text{'}}=0.0833\text{N}$