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3 years ago

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A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.150 A. The loop is inside a solenoid,

with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A.
(a) Find the force on each side of the loop.
magnitude = µN.

(b) Find the magnitude of the torque acting on the loop.
= N · m

1 answer:

Bond [772]3 years ago

6 0

Answer:

a

The force acting on one of the sides the square loop is 169.6 N

b

The torque acting on the loop is ZERO

Explanation:

From the question we are given that

permeability of free space $\mu_0$ $= 4\pi*10^{-7} T\cdot m/A$

Number of turns of the solenoid per unit length , n = 30 turns per cm

= 3000 turns per m

The current flowing through the solenoid $I_1$ = 15A

Length of the side of the loop, L = 2 cm

= 2 cm [ $\frac{1*10^{-2}m}{1cm}$ ]

= $2*10^{-2} m$

Angle , $\theta =90$ °

The current flowing through the wire $I_2 = 0.15A$

Generally The formula for Magnetic field inside of a solenoid is

$B = \mu_0nI_1$

Where

$\mu_0$ is the permeability of free space

n is the number of turns of the solenoid per unit length

$I_1$ is the current through the solenoid

Also the formula for force acting on one of the sides of a square loop is

$F = BI_2Lsin \theta$

Where $I_2$ is the current through the wire ,

L is the length of the sides of the loop,and

$\theta$ is the angle between B and $I_2$

Substituting the given values into the first formula

$B = (4 \pi *10^{-7})(3000)(15)$

$= 5.65*10^{-2} T$

To obtain the force acting on the loop by substituting the appropriate value into the second formula

$F = (5.65*10^{-2}(0.15)(2*10^{-2}(sin90)))$

$169.6 \mu N$

The torque acting on the lop is zero because the

The forces on the loop are acting in the plane of the loop

These force are perpendicular to the sides

and their direction is away from the inner side of the loop

the only effect the orientation of these forces can cause is stretching of the loop

This way by which the forces are oriented cannot cause any rotation

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1 year ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

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Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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3 years ago