Question

problem 20.01
what is the magnitude of the force per meter of length on a straight wire carrying an 8.10 - a current when perpendicular to a 0.60 - t uniform magnetic field?
express your answer to two significant figures and include the appropriate units.
part b
what if the angle between the wire and field is 47.0°?
express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall the formula for force per unit - length when wire is perpendicular to magnetic - field

The formula for the force per unit length on a current - carrying wire in a magnetic field is $\frac{F}{l}=IB\sin\theta$. When the wire is perpendicular to the magnetic field, $\theta = 90^{\circ}$ and $\sin\theta=1$. Given $I = 8.10\ A$ and $B = 0.60\ T$.
$\frac{F}{l}=IB\sin90^{\circ}$

Step2: Calculate the force per unit length

Substitute the values of $I$ and $B$ into the formula:
$\frac{F}{l}=8.10\ A\times0.60\ T\times1 = 4.86\ N/m$
Rounding to two significant figures, $\frac{F}{l}=4.9\ N/m$

Step3: For the case when $\theta = 47.0^{\circ}$

Use the formula $\frac{F}{l}=IB\sin\theta$. Substitute $I = 8.10\ A$, $B = 0.60\ T$, and $\theta = 47.0^{\circ}$.
$\sin47.0^{\circ}\approx0.731$
$\frac{F}{l}=8.10\ A\times0.60\ T\times0.731$
$\frac{F}{l}=8.10\times0.60\times0.731\ N/m=3.56\ N/m$
Rounding to two significant figures, $\frac{F}{l}=3.6\ N/m$

Answer:

Part A: 4.9 N/m
Part B: 3.6 N/m