Question

A 1.00 m long wire carrying a0.729 A current is oriented at38.0 degrees to a magnetic field,and feels a magnetic force of0.0918 N. How strong is themagnetic field?(Unit = T)

Answer 1

Given:

• Length of wire, L = 1.00 m

,

• Current, I = 0.729 A

,

• θ = 38.0 degrees

,

• Magnetic force, F = 0.0918 N

Let's find the strength of the magnetic field.

To find the strength of the magnetic field, apply the formula:

$F=ILBsin\theta$

Where:

B is the strength of the field.

Rewrite the formula for B, plug in the values and evaluate.

We have:

$\begin{gathered} B=\frac{F}{ILsin\theta} \\ \\ B=\frac{0.0918}{0.729*1.00*sin38} \\ \\ B=\frac{0.0918}{0.448817} \\ \\ B=0.2045\text{ T} \end{gathered}$

Therefore, the strength of the magnetic field is 0.2045 T.

• ANSWER:

0.2045 T