Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ =
...................(1)
Where:
.
{ B is magnetic field }
{A is cross-sectional area}
.
No. of turns in coil.
.
Rate change of induced Emf.
Here,
Considering the case :-
&
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =
∈1 =
∈1 =![4 [-N\times\frac{d\phi}{dt}]](https://tex.z-dn.net/?f=4%20%5B-N%5Ctimes%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5D)
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.
Answer:
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune, Milky Way ;-;
Answer:

Explanation:
The radius of the circumference described by a particle in a cyclotron is given by:

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.
Kinetic energy is defined as:

Solving this for mv:

Replacing (2) in (1):

For protons, we have:

For alpha particles, we have:

Dividing (4) in (3):

The frequency of an oscillation is equal to the reciprocal of the period:

where
f is the frequency
T is the period
In our problem, the period is T=8.01 s, therefore the corresponding frequency is
Answer:
0.0170m or 1.70cm
Explanation:
L=(μ₀*N^2*A)/(l)
N=x/2*pi*r so r=x/2*pi*N
A=pi*r^2=(pi*x^2)/(4*pi^2*N^2)
L=(μ₀*N^2*A)/(l)=(μ₀*N^2*pi*x^2)/(4*pi^2*N^2*l)=(μ₀*x^2)/(4*pi*l)
l=(μ₀*x^2)/4*pi*L)=[(4*10^-7)*(1.15m)/4*pi*2.47*10^-6H)=0.0170m or 1.70cm
Hope this helps. Any questions please feel free to ask. Thanks!