Lift the Load!

Electricity and Magnetism Level 5

Consider a loop of freely deformable conducting wire with insulation wire of length $2l$ , the two ends of which are fixed (permanently) to the ceiling. A load of mass $m$ is fixed to the middle of the wire (the mass of which is negligible). There is also a horizontal magnetic field of induction $B$ ; free-fall acceleration is $g$ . A current $I$ is passed through the wire. Neglect the field induced by the wires.

The maximum height by which the load can be lifted is $l(1-x)$ .
To lift the load by $l\left(1-\frac{3}{\pi}\right),$ the current in the wire is $\frac{mgy}{lB}$ .

Find $x+\frac{\sqrt{3}y}{\sqrt{2}}$

The answer is 1.3771.

1 solution

Rajdeep Brahma
Jul 7, 2018

Since the tension in wire is uniform so both halves of the wire will take the form of a circular segment.

$mg=2Tcos\alpha$

$R=\frac{l}{2\alpha}$ = $\frac{T}{IB}$ and the lifting height $h=l-2Rsin\alpha=l(1-\frac{sin\alpha}{\alpha})$

Now you should be able to conclude that $x=\frac{2}{\pi}$ and $y=\frac{\pi}{3*\sqrt{3}}$ ( $\alpha=30$ )

Try to prove that $sinx>=\frac{2x}{\pi}$ in the range $(0,\frac{\pi}{2})$ by drawing the graph $y=sinx$ and $y=\frac{2x}{\pi}$ .

there is something wrong with your solution. mg should be equal to 2 T cos alpha. im actually getting alpha to be exactly equal to 30 degrees. try solving it again. the answer should be x = 2 / pi, and y = sqrt (3) * pi / 9.

Ramon Vicente Marquez - 2 years, 11 months ago

yup $mg=2Tcos\alpha$ ...that was a typo but I am getting the same answer sir.You are welcome to point out my mistake.

rajdeep brahma - 2 years, 11 months ago

consider the left half of the wire. the net magnetic force acting on that portion of the wire is I (3 L / pi) B to the left. it can be shown that, in a uniform magnetic field, the net magnetic force on any curved wire from point A to point B is equivalent to the net magnetic force on a straight wire from point A to point B. how did you evaluate the forces acting on the wire?

Ramon Vicente Marquez - 2 years, 11 months ago

Yes exactly that is what I got Also this was a pathfinder question With the same answer given as Ramon Sir's

Suhas Sheikh - 2 years, 10 months ago

Hey this is. Kinda bad You changed the question Instead of considering your inaccuracy :(

Suhas Sheikh - 2 years, 10 months ago

what can I do else?i can not change the answer

rajdeep brahma - 2 years, 10 months ago

Lift the Load!

The answer is 1.3771.

1 solution

m g = 2 T c o s α mg=2Tcos\alpha m g = 2 T c o s α

R = l 2 α R=\frac{l}{2\alpha} R = 2 α l ​ = T I B \frac{T}{IB} I B T ​ and the lifting height h = l − 2 R s i n α = l ( 1 − s i n α α ) h=l-2Rsin\alpha=l(1-\frac{sin\alpha}{\alpha}) h = l − 2 R s i n α = l ( 1 − α s i n α ​ )

Now you should be able to conclude that x = 2 π x=\frac{2}{\pi} x = π 2 ​ and y = π 3 ∗ 3 y=\frac{\pi}{3*\sqrt{3}} y = 3 ∗ 3 ​ π ​ ( α = 30 \alpha=30 α = 3 0 )

Try to prove that s i n x > = 2 x π sinx>=\frac{2x}{\pi} s i n x > = π 2 x ​ in the range ( 0 , π 2 ) (0,\frac{\pi}{2}) ( 0 , 2 π ​ ) by drawing the graph y = s i n x y=sinx y = s i n x and y = 2 x π y=\frac{2x}{\pi} y = π 2 x ​ .

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$mg=2Tcos\alpha$

$R=\frac{l}{2\alpha}$ = $\frac{T}{IB}$ and the lifting height $h=l-2Rsin\alpha=l(1-\frac{sin\alpha}{\alpha})$

Now you should be able to conclude that $x=\frac{2}{\pi}$ and $y=\frac{\pi}{3*\sqrt{3}}$ ( $\alpha=30$ )

Try to prove that $sinx>=\frac{2x}{\pi}$ in the range $(0,\frac{\pi}{2})$ by drawing the graph $y=sinx$ and $y=\frac{2x}{\pi}$ .