As the magnet is moved inside a coil of wire, the number of lines of magnetic field passing through the coil changes. Faraday stated that : it is the change in the number of field lines passing through the the coil of wire that induces emf in the loop. Specifically, it is the rate of change in the number of magnetic field lines passing through the loop that determines the induced emf. There is a term called magnetic flux same as electric flux, this magnetic flux can be a measure of the number of field lines passing through a surface. It is given by ( Φ=ΣB. dA. Where B is magnetic field and dA is small elementary area). The induced emf is given by (ξ = dΦ/dt). This equation states that THE MAGNITUDE OF THE INDUCED CURRENT IN A CIRCUIT IS EQUAL TO THE RATE AT WHICH THE MAGNETIC FLUX THROUGH THE CIRCUIT IS CHANGING WITH TIME. So more rapid you move the coil, more will be the change in flux and hence more emf will be produced. So option D is the correct answer. I hope this long description will help you out.
Answer: The potential difference from the sphere's surface to its center is zero.
Explanation:
The expression for the potential difference due to charge is as follows;

Here,
is the potential difference, q is the charge, r is the position and
is the absolute permittivity of free space.
In the given problem, a solid sphere of radius r carries charge q distributed uniformly throughout its volume.
To find the potential difference from the sphere's surface to its center. Put r=0 in the expression of the potential difference.


Therefore, the potential difference from the sphere's surface to its center is zero.
Answer:
15 deg
Explanation:
Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .
Given:
For first ball, θ1 = 75◦
initial velocity for both the balls, u = 27.2 m/s
for second ball, θ2 = ?
since distance covered by both the balls is same.
Therefore,..
R1=(u^{2} sin2\alpha _{1}) /g[/tex]
the range for the first ball
the range for the second ball
R2=(u^{2} sin2\alpha _{2}) /g[/tex]
(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]
sin2\alpha _{2})=sin2\alpha _{1})
=sin^-1(sin2\alpha _{1})
=1/2sin^-1(sin2\alpha _{1})
=
15 deg
Answer:
what did u say and what language are you speaking in