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padilas [110]
3 years ago
12

A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

ular to the strip. A current i = 32 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V.
Physics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current  i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28  × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
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Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

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∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

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F= 282 N

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g = 9.8 m/s² : acceleration due to gravity

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Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

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∑Fx = m*ax    , ax=0

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a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

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We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

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