The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.
Self-Inductance: -
The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.
We know that the self-inductance of the coil is denoted by: -
L= µ *π*(r)^2*(N)^2*l
Where
L= Self-Inductance of the coil
µ= Magnetic Permeability Constant
r= Radius of the coil
l= Length of the coil
N= Number of turns of the coil
Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.
So,
On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.
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Answer:
doubled the initial value
Explanation:
Let the area of plates be A and the separation between them is d.
Let V be the potential difference of the battery.
The energy stored in the capacitor is given by
U = Q^2/2C ...(1)
Now the battery is disconnected, it means the charge is constant.
the separation between the plates is doubled.
The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.
C' = C/2
the new energy stored
U' = Q^2 / 2C'
U' = Q^2/C = 2 U
The energy stored in the capacitor is doubled the initial amount.
When your hand hits the table the table will vibrate and your hand will be numb for two to three seconds
Answer:
Explanation:
given,
Radius of the solid rod, R = 5.3 cm
Electric field strength,E = 22 kN/C
Let the volume charge density be ρ
From Gauss law
ε₀ is the permitivity of free space
R is the radius of the rod
and also,
ρ is the volume charge density
Hence, the volume charge density is equal to