Answer:
Induced emf, 
Explanation:
We have,
Number of turns in the coil, N = 40
Radius of coil, r = 3 cm = 0.03 m
The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.
It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

, is magnetic flux

So, the magnitude of induced emf is
.
Answer:
D
Explanation:
The law of reflection states that:
- The incident ray and the reflected ray lie on the same plane
- The angle of reflection is equal to the angle of incidence
The angle of incidence is the angle between the direction of the incident ray and the normal to the surface, while the angle of reflection is the angle between the direction of the reflected ray and the normal to the surface.
From the figure, we see that the only situation where the angle of reflection is equal to the angle of incidence is ray D.
<span>Uniform. A parallel plate capacitor is a simple arrangement of electrodes and dielectric to form a capacitor. Here two parallel conductive plates are used as electrodes with a medium or dielectric in between them. Charge separation in a parallel-plate capacitor causes an internal electric field, which is uniform.</span>
I = V/Z
V = voltage, I = current, Z = impedance
First let's find the total impedance of the circuit.
The impedance of the resistor is:
= R
R = resistance
Given values:
R = 1200Ω
Plug in:
= 1200Ω
The impedance of the inductor is:
= j2πfL
f = source frequency, L = inductance
Given values:
f = 59Hz, L = 2.4H
Plug in:
= j2π(59)(2.4) = j889.7Ω
Add up the individual impedances to get the Z, and convert Z to polar form:
Z =
+ 
Z = 1200 + j889.7
Z = 1494∠36.55°Ω
I = V/Z
Given values:
V = 170∠0°V (assume 0 initial phase)
Z = 1494∠36.55°Ω
I = 170∠0°/1494∠36.55°Ω
I = 0.1138∠-36.55°A
Round the magnitude of I to 2 significant figures and now you have your maximum current:
I = 0.11A