Answer
a. 5.8 N/m
b. 3.3 N/m
Work Step by Step
a. Use equation 20–1 to calculate the force with an angle of 90 degrees, and a length of 1 meter.
$$F=I\mathcal{l}Bsin\theta$$
$$\frac{F}{\mathcal{l}}=IBsin\theta=(6.40A)(0.90T)sin90^{\circ}=5.8N/m$$
b. Change the angle.
$$\frac{F}{\mathcal{l}}=IBsin\theta=(6.40A)(0.90T)sin35^{\circ}=3.3N/m$$