Answer:
236.3 x
C
Explanation:
Given:
B(0)=1.60T and B(t)=-1.60T
No. of turns 'N' =100
cross-sectional area 'A'= 1.2 x
m²
Resistance 'R'= 1.3Ω
According to Faraday's law, the induced emf is given by,
ℰ=-NdΦ/dt
The current given by resistance and induced emf as
I = ℰ/R
I= -NdΦ/dtR
By converting the current to differential form(the time derivative of charge), we get
= -NdΦ/dtR
dq= -N dΦ/R
The change in the flux dФ =Ф(t)-Ф(0)
therefore, dq =
(Ф(0)-Ф(t))
Also, flux is equal to the magnetic field multiplied with the area of the coil
dq = NA(B(0)-B(t))/R
dq= (100)(1.2 x
)(1.6+1.6)/1.3
dq= 236.3 x
C
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁

where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
A, something to remember is that if the numbers are even the answer too will be even. Every time hope this helps :>