91Ó°ÊÓ

• The given data can be listed below as

• The magnetic field is,${\mathrm{B}}_{\mathrm{x}}=-0.242\mathrm{T}$ .

• The magnetic field is, ${\mathrm{B}}_{\mathrm{y}}=-0.985\mathrm{T}$.

• The magnetic field is, ${\mathrm{B}}_{\mathrm{z}}=-0.336\mathrm{T}$.

• The value of current is, I = 7.40 A.

• The length of bar is, l = 25.0 cm .

The term magnetic field may be defined as the area around the magnet behaves like a magnet.

According to the question we can write as $\stackrel{→}{l}=l\hat{k}$, the magnetic force acting on the wire is

$\begin{array}{r}\stackrel{→}{\mathrm{F}}=\mathrm{I}\stackrel{→}{\mathrm{I}}×\stackrel{→}{\mathrm{B}}\\ \stackrel{→}{\mathrm{F}}=\mathrm{I}\left(\mathrm{I}\widehat{\mathrm{k}}\right)×\left({\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{k}}\right)\\ \stackrel{→}{\mathrm{F}}=\mathrm{II}\left[\left(−{\mathrm{B}}_{\mathrm{y}}\right)\widehat{\mathrm{i}}+\left({\mathrm{B}}_{\mathrm{x}}\right)\widehat{\mathrm{j}}\right]\\ \mathrm{So}\\ {\mathrm{F}}_{\mathrm{x}}=−{\mathrm{IIB}}_{\mathrm{y}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{r}{\mathrm{F}}_{\mathrm{x}}=−(7.40\mathrm{A})(0.250\mathrm{m})(−0.985\mathrm{T})\\ {\mathrm{F}}_{\mathrm{x}}=1.82\mathrm{N}\\ \mathrm{And}\\ {\mathrm{F}}_{\mathrm{y}}=−{\mathrm{IIB}}_{\mathrm{x}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{r}{F}_y=−\left(7.40\mathrm{A}\right)\left(0.250\mathrm{m}\right)(−0.242\mathrm{T})\\ F_y=0.448\mathrm{N}\end{array}$

And the force in the z-direction zero.

Hence, the components of the magnetic force on the wire are $F_x=1.82N$and localid="1668239074264" $F_y=-0.448N$.

The magnitude of the net magnetic force on the wire can be calculated as

$F=\sqrt{F_x^2+F_y^2}$

Substitute all the value in the above equation.

$\begin{array}{r}F=\sqrt{F_x^2+F_y^2}\\ F=\sqrt{(1.82N{)}^2+(0.448N{)}^2}\\ F=1.88N\end{array}$

Hence, the magnitude of the net magnetic force on the wire is 1.88 N.