A stiff wire 50.0 cm long is 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

Question

A stiff wire 50.0 cm long is 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. A current of 20.0 A flows in the wire — down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field B = (.318 i)T. Determine the magnitude and the direction of the total force on the wire.

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RuslanHeatt 2 months 2021-07-30T02:38:02+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-30T02:39:17+00:00

    Answer:

    F=2.38i – 4.57j

    magnitude of F=5.15N

    Explanation:

    The force on the wire is given by

    F=Il \ X B

    in this problem we have to compute the force for both sections if the wire. Hence we have

    F=Il_1\ X B+Il_2\ X B\\\\

    The direction of l1 is k. The direction of l2 is obtained by using the slope of the line y=2x

    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}

    By applying the cross product we have

    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}

    and its magnitude is

    |F|=\sqrt{(2.38)^2+(4.57)^2}=5.15N

    HOPE THIS HELPS!!

    0
    2021-07-30T02:39:55+00:00

    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

    \frac{y}{x} = 2

    tan θ = 2

    Therefore,

    slope m = tan θ = y / x

    \theta=\tan^-^1(2)=63.4^0

    Then length of each section is 25.0cm

    so, length vector of the wire is

    \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

    And magnetic field is B = (0.318T)i

    Therefore,

    \bar F = \hat I (\bar l \times \bar B)

    \bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i]

    = (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

    Magnitude of the force is

    F = \sqrt{(-1.59N)^2+(-1.425N)^2\\} \\F = 2.135N

    Direction is

    \alpha = \tan^-^1(\frac{-1.425N}{-1.59N} )\\\\= 41.8^0

    Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

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