Answer:
Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis
Explanation:
The stiff wire 50.0cm long bent at a right angle in the middle
One section lies along the z axis and the other is along the line y=2x in the xy plane
[tex]\frac{y}{x} = 2[/tex]
tan θ = 2
Therefore,
slope m = tan θ = y / x
[tex]\theta=\tan^-^1(2)=63.4^0[/tex]
Then length of each section is 25.0cm
so, length vector of the wire is
[tex]\hat I= (-25.0)\hat k +(25.0) \cos 63.4^0 \hati +(25.0) \sin63.4^0 \hatJ\\\\\hat I = (11.2) \hat i + (22.4) \hat j - (23.0) \hat k[/tex]
And magnetic field is B = (0.318T)i
Therefore,
[tex]\bar F = \hat I (\bar l \times \bar B)[/tex]
[tex]\bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i][/tex]
[tex]= (20.0)(i(0)+j(-0.250)(0.318T)+k[0-(0.224m)(0.318T)]\\\\=(20.0)(-0.250)(0.318)j-(20.0)(0.224)(0.318T)\\\\=-(1.59N)j-(1.425N)k[/tex]
Magnitude of the force is
[tex]F = \sqrt{(-1.59N)^2+(-1.425N)^2\\} \\F = 2.135N[/tex]
Direction is
[tex]\alpha = \tan^-^1(\frac{-1.425N}{-1.59N} )\\\\= 41.8^0[/tex]
Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis
Answer:
F=2.38i - 4.57j
magnitude of F=5.15N
Explanation:
The force on the wire is given by
[tex]F=Il \ X B[/tex]
in this problem we have to compute the force for both sections if the wire. Hence we have
[tex]F=Il_1\ X B+Il_2\ X B\\\\[/tex]
The direction of l1 is k. The direction of l2 is obtained by using the slope of the line y=2x
[tex]tan\theta =2\\\theta=63.43\°\\l_2=(0.5cos(63.43))\hat{i}+(0.5sin(63.43))\hat{j}=0.22\hat{i}+0.89\hat{j}[/tex]
By applying the cross product we have
[tex]F=-(20.0A)(0.5m)(0.318T)\hat{j}+(20.0A)(0.445m)(0.318T)\hat{i}-(20.0A)(0.22m)(0.318T)\hat{j}\\\\F=-3.18N\hat{j}+2.83N\hat{i}-1.39N\hat{j}=2.38N\hat{i}-4,57N\hat{j}[/tex]
and its magnitude is
[tex]|F|=\sqrt{(2.38)^2+(4.57)^2}=5.15N[/tex]
HOPE THIS HELPS!!
