Answer:
Current in outer circle will be 15.826 A
Explanation:
We have given number of turns in inner coil 
Radius of inner circle 
Current in the inner circle 
Number of turns in outer circle 
Radius of outer circle 
We have to find the current in outer circle so that net magnetic field will zero
For net magnetic field current must be in opposite direction as in inner circle
We know that magnetic field is given due to circular coil is given by
For net magnetic field zero
So 
Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive
Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:
Replacing the dat we obtain F=82 nN.
The force is repulsive because the points charged have the same sign.
Answer: A flower pot falling
Explanation:
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
