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Sonbull [250]
3 years ago
5

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

ch of the wires, the net field is zero here? 2. Note the distance between the field lines from one of the wires. Are they equally space? Explain. 3. With currents of different magnitudes passing through the two wires, how do the forces on the two wires compare? What is the reason for this?

Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

Explanation:

1 )  Magnetic field due to long current carrying wire : B = \frac{U_{0} I}{2\pi d}

Therefore the net magnetic field due the both wires ; B = B_{1} + B_{2} . when we adjust the current I_{1} = I_{2} then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

F = \frac{U_{0}I_{1}I_{2}   }{2\pi d }

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D

The closer the field lines, the stronger the field, whereas for electric field, the higher the electric flux passing through, the stronger the electric field.

Explanation:

Another difference between the two is that magnetic fields formed closed loops around the magnetic while electric fields do not. Electric fields are measured in two dimensional while magnetic fields in 3 dimensional. The electric field line can do work while magnetic fields can not (because particles in magnetic field remain constant even though the charge may change direction).

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4 years ago
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The force required to maintain an object at a constant velocity in free space is equal to
alex41 [277]

Answer:

The force required to maintain an object at a constant velocity in free space is equal to Zero.

7 0
3 years ago
A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is
Studentka2010 [4]

Answer:-81.29 J

Explanation:

Given

mass of ball m=2.35 kg

Length of string L=3.53 m

height of Room h=5.03 m

Gravitational Potential Energy is given by

P.E.=mgh

where h=distance between datum and object

here Reference is ceiling

therefore h=-3.53 m

Potential Energy of ball w.r.t ceiling

P.E.=2.35\times 9.8\times (-3.53)=-81.29 J

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling    

3 0
3 years ago
A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi
Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft \omega _1=420\ rev/min

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment  of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

L_1=L_2

I_1\omega _1=I_2\omega _2

I\times (420)=7 I\times (\omega _2)

\omega _2=\frac{420}{7}

\omega _2=60\ rev/min  

8 0
4 years ago
An object that is initially at rest on a frictionless floor is struck by a flying rock with an impulse jfonox. what is the final
jasenka [17]

Answer:

jfonox

Explanation:

since impulse is the change in momentum given by

Impulse = Force x time = change in momentum

I = ft = mv-mu= m(v-u)= jfonox

since the object is initially at rest

u=0,

mv is the final momentum of the object

I =Ft = m(v - 0) = jfonox

mv= jfonox

this show that the final momentum of the object at rest is equal to the impulse received

if  any of the mass m, final velocity v or the impact force and the duration of impact of the object is known then we can find or quantify the final momentum.

5 0
3 years ago
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