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3 years ago

I need help with this someone help me?

Physics

1 answer:

Brilliant_brown [7]3 years ago

3 0

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

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Given that,

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$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

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$F_{130}=0$

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The direction of force is out of page.

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The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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