# Question #0d4e0

Sep 18, 2017

$F = 22.5$ $\text{N}$

#### Explanation:

Let's use the equation $F = B I L \sin \left(\theta\right)$; where $B$ is the magnitude of the magnetic field, $I$ is the current carried by the wire, $L$ is the length of the wire, and $\theta$ is the angle the wire makes with the magnetic field:

$R i g h t a r r o w F = 1.24$ $\text{T} \cdot 12.8$ $\text{A} \cdot 1.99$ $\text{m} \cdot \sin \left({45.4}^{\circ}\right)$

$R i g h t a r r o w F = 1.24$ ${\text{N" cdot "s" cdot "C"^(- 1) cdot "m}}^{- 1} \cdot 12.8$ $\text{A} \cdot 1.99$ $\text{m} \cdot 0.71202604599$

$R i g h t a r r o w F = 1.24$ ${\text{N" cdot "s" cdot "A"^(- 1) cdot "s"^(- 1) cdot "m}}^{- 1} \cdot 25.472$ $\text{A" cdot "m} \cdot 0.71202604599$

$R i g h t a r r o w F = 31.58528$ $\text{N} \cdot 0.71202604599$

$R i g h t a r r o w F = 22.48954203$ $\text{N}$

$\therefore F = 22.5$ $\text{N}$

Therefore, the magnetic force on the wire is around $22.5$ $\text{N}$