Question

A long wire carrying 4.50 A of current makes two 90° bends, as shown in Fig. E27.35. The bent part of the wire passes through a uniform 0.240-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

Short answer

The magnitude of a long wire carrying current of 4.5 A and making a bend of $90°$

is 0.724N and the angle is$63.4°$

Step-by-step solution

Horizontal force

The force acting on the first and the third segment are all in the same direction that is downwards acting on the second segment which points to the right direction, Between the magnetic field and the length is , now using:

$F=llB\sin \varphi$

we can write the vertical forces as:

$\begin{array}{r}{\mathrm{F}}_{\mathrm{y}}={\mathrm{IL}}_{\mathrm{a}}\mathrm{B}+{\mathrm{IL}}_{\mathrm{c}}\mathrm{B}\\ =(4.50)\times (0.240)+(4.50\mathrm{A})(60.0\mathrm{cm}-\mathrm{x})(0.240\mathrm{T})\\ =(4.50\mathrm{A})(0.60\mathrm{mm})(0.240\mathrm{T})\\ =0.648\mathrm{N}\\ \mathrm{Now},\mathrm{the}\mathrm{horizontal}\mathrm{force}\mathrm{is}:\\ {\mathrm{F}}_{\mathrm{x}}=\mathrm{ILB}\\ =(4.50\mathrm{A})(0.30\mathrm{m})(0.240\mathrm{T})\\ =0.324\mathrm{T}\end{array}$

Now, the horizontal force is:

Calculating the force and the and the angle

$\begin{array}{r}\tan \mathrm{\theta }=\frac{{\mathrm{F}}_{\mathrm{y}}}{{\mathrm{F}}_{\mathrm{x}}}\\ =\frac{0.648}{0.324}\\ =2.00\end{array}$

Therefore, the angle of the wire is:

$\begin{array}{r}\theta =2.00\\ ={63.4}^{\circ }\end{array}$