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olga55 [171]
3 years ago
5

We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpe

ndicular to the direction of the field. The loop is moving with a constant velocity of 2.1 m/s into the field region starting from a point outside the field. If the loop has a resistance of 3 Ω, find the force on the loop as it just enters the magnetic field region. (Enter the magnitude.)
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

Force will be F=8.33\times 10^{-4}N

Explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by i=\frac{Bvd}{R}=\frac{1.5\times 2.1\times 0.023}{3}=0.02415A

Power is given by P=i^2R=0.02415^2\times 3=0.00174watt

We know that power is also given by P=Fv

So 0.00174=F\times 2.1

F=8.33\times 10^{-4}N

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3 years ago
A worker wants to turn over a uniform 1040-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
zmey [24]

> How hard does the floor push on the crate?

We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1040N. 
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Fpn = Fp * cos(θp) 
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= 573.23 N 

The total normal force Fn then is: 
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= 1040 + 573.23
= 1613.2 N

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: </span>

Fh = Fph + Fgh 
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> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 760.7 N / 1613.2 N

= 0.472

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3 years ago
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