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Reptile [31]
4 years ago
14

A coil with magnetic moment 1.48 Aâ‹…m2 is oriented initially with its magnetic moment antiparallel to a uniform magnetic field

of magnitude 0.805 T .What is the change in potential energy of the coil when it is rotated 180 degrees, so that its magnetic moment is parallel to the field? Express the answer in joules.
Physics
1 answer:
serg [7]4 years ago
5 0

Answer:

-2.3828 J

Explanation:

Magnetic moment = 1.48 Am²

B = Magnetic field = 0.805 T

\theta_i = Initial turns = 180°

\theta_f = Final turns = 360°

Initial in potential energy

U_i=-mBcos\theta_i\\\Rightarrow U_i=-1.48\times 0.805\times cos180\\\Rightarrow U_i=1.1914\ J

Final in potential energy

U_f=-mBcos\theta_f\\\Rightarrow U_f=-1.48\times 0.805\times cos360\\\Rightarrow U_f=-1.1914\ J

Change potential energy would be

\Delta U=U_f-U_i\\\Rightarrow \Delta U=-1.1914-1.1914\\\Rightarrow \Delta U=-2.3828\ J

Change in potential energy is -2.3828 J

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(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

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v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

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v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

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v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

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As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

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