Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
length of each rod is 1.1 m, while the mass of each is 0.11 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.3 mm. Determine the current in the rods.
1 answer:
Answer:
I = 215.76 A
Explanation:
The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.
Fm = u_o*I^2*L/2*π*d
where I is the current in each rod, d = 0.0082 m is the distance 27rId
between each, L = 0.85 m is the length of each rod.
Fm = 4π*10^-7*I^2*1.1/2*π*0.0083
The mass of each rod is m = 0.0276 kg
F_m = mg
4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8
I = 215.76 A
note:
mathematical calculation maybe wrong or having little bit error but the method is perfectly fine
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Answer:
Explanation:
Consider two particles are initially at rest.
Therefore,
the kinetic energy of the particles is zero.
That initial K.E. = 0
The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are
now, since both the masses have mass m
therefore,
= m/2
The final K.E. of the particles is
Distance between two particles is d and the gravitational potential energy between them is given by
By law of conservation of energy we have
Now plugging the values we get
This the required relation between G,m and d