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2 years ago

When is the force on a current-carrying wire in a magnetic field at its strongest?

1 answer:

4 0

The forces on a current-carrying wire in a magnetic field are at their strongest when the current is at a 90-degree angle to the field. Option D is correct.

<h3>What is a magnetic field?</h3>

It is the type of field where the magnetic force is obtained. The magnetic force is obtained by the field felt around a moving electric charge.

The complete question is;

"When is the force on a current-carrying wire in a magnetic field at its strongest?

-when the current is at a 0-degree angle to the field

-when the current is at a 30-degree angle to the field

-when the current is at a 45-degree angle to the field

-when the current is at a 90-degree angle to the field"

The magnetic force is found as;

F=BILSINΘ

Where,

Magnetic Field, B

Length of the wire, L

The angle between field and current, Θ

When Θ=90°

The value of the magnetic force is;

F=BIL

When the current is flowing at a 90-degree angle to the magnetic field, the forces acting on a wire carrying a current are the strongest.

Hence, option D is correct.

To learn more about the magnetic field, refer to the link;

brainly.com/question/19542022

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin

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Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

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$W = \frac{1}{2}k_{eq}x^2$

Making the spring constant $k_{eq}$ the subject of the formula; we have:

$k_{eq} = \frac{2W}{x^2}$

Substituting our given values, we have:

$k_{eq} = \frac{2*80}{0.180^2}$

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The magnitude of the force that must be apply to the hold platform in this position is given by the formula :

$F = k_{eq}x$

$F = 4938.27*0.180$

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3 years ago

If you wanted to calculate the density of a baseball and you knew its mass and volume you would need the density formula and a(n

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3 years ago

Read 2 more answers

A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the building.

Nadusha1986 [10]

The height of the building is 53.4 m (projectile motion)

Explanation:

The motion of the ball is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

First of all, we consider the horizontal motion, which is a uniform motion at constant speed. The equation that gives the horizontal distance travelled is:

$d=v_x t$

where:

$v_x = 50 m/s$ is the horizontal velocity of the ball, which is constant since there are no forces in this direction

d = 165 m is the horizontal distance travelled by the ball

Solving for t, we find the time of flight:

$t=\frac{d}{v_x}=\frac{165}{50}=3.3 s$

Now we analyze the vertical motion: it is an accelerated motion, we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s is the vertical displacement, which is the height of the building

u = 0 is the initial vertical velocity of the ball

t = 3.3 s is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for s, we find the height of the building:

$s=0+\frac{1}{2}(9.8)(3.3)^2=53.4 m$

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4 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji

WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is $-14.75^{o}$

Explanation:

Force by Jack in vector form

$\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)$

Force by Jill in Vector form is given by

$\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

Force by Jane is

$\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}$

Net force is:

$\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}$

Hence

$\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

The net force will be given by

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}$

The direction of net force is:

$\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}$

The angle with East direction is $-14.75^{o}$

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is $-14.75^{o}$ and it is $14.75^{o}$ from the south.

5 0

3 years ago