Chapter 20, Problem 1P

To determine

Part (a) To Determine:

Force per metre length of straight wire carrying current 6.40 A placed perpendicular to Uniform magnetic field of 0.90 T.

Answer to Problem 1P

Solution:

Force per metre length of straight wire = 5.76 N/m

Explanation of Solution

Given:

Current I = 6.40 A; Magnetic field B = 0.90 T and Ϙ = 90^o

Formula Used:

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

Calculation:

In order to find the value of force per metre length of current carrying conductor, we have to use the equation,

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

For force per unit length equation becomes,

Force per unit length = $IB\mathrm{Sin}\theta$

Putting the values in this equation we get,

Force per unit length = $6.40\times 0.90\times \mathrm{Sin}90$

= $6.40\times 0.90\times 1$

= $5.76\text{N/m}$

To determine

Part (b) To Determine:

Force per metre length of straight wire carrying current 6.40A placed 35⁰ to Uniform magnetic field of 0.90T

Answer to Problem 1P

Solution:

Force per metre length of straight wire = 3.28 N/m

Explanation of Solution

Given:

Current I=6.40A; Magnetic field B= 0.90 T and Ϙ=35⁰

Formula Used:

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

Calculation:

To find the value of force per metre length of current carrying conductor, we have to use the equation,

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

For force per unit length equation becomes

Force per unit length becomes = $IB\mathrm{Sin}\theta$

$=(6.40)(0.90)\mathrm{Sin}{35}^0$

$=(6.40)(0.90)(0.57)$

$=3.28\text{N/m}$