Answer:
2 Joules
Explanation:
Kinetic Energy = 1/2 of (Mass x Velocity^2)
Answer:
67.44 V
Explanation:
Number of turns N =123
Radius = 2.41 cm =0.0241 m
The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T
Time interval dt = 0.151 sec
We know that the induced emf 
Area 
Putting all these values in emf equation
here negative sign indicates that it opposes the cause due to which it is produced
Answer:
F = 0
Explanation:
The magnetic force is described by two expressions
for a moving charge
F = q v x B
for a wire with a current
F = I L xB
bold indicates vectors
let's write this equation in module form
F = I L B sin θ
where the angle is between the direction of the current and the direction of the magnetic field
In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current
The magnetic field of the Earth goes from the south to the north and in this part it is horizontal
Therefore the current and the magnetic field are parallel, the angle between them is zero
sin 0 = 0
consequently the magnetic force is zero
F = 0
Answer:
0.23 s
Explanation:
First of all, let's find the time constant of the circuit:

where
is the resistance
is the capacitance
Substituting,

The charge on a charging capacitor is given by
(1)
where
is the full charge
we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Substituting this into eq.(1) we find
