Magnetic field 'B' at a distance 'r' from an substantially
large conductor carrying current 'i' = (2x10^ -7)('i' ) / r
Magnetic field 'B' beyond the wire= (2x10^ -7)(8.15x10^18x1.6x10^ - 19 ) /0.046
=5.7 x10^ -12 tesla
As electrons move from west to east, the conventional current is from east to
west.
By means of Maxwell's right handed corkscrew rule, the way of magnetic field is
from south to north.
Answer:
leg D → leg C → leg E → leg A →leg B
Explanation:
PLEASE VOTE ME BRAINLYIEST!!!!!!!!
Answer:
81.3ohms
Explanation:
Resistance is known to provide opposition to the flow of electric current in an electric circuit.
Power dissipated by the computer is expressed as;
Power = current (I) × Voltage(V)
P = IV... (1)
Note that from ohms law, V = IR
I = V/R ... (2)
Substituting equation 2 into 1, we will have;
P = (V/R)×V
P = V²/R.. (3)
Given source voltage = 100V, Power dissipated = 123W
To get resistance R of the computer, we will substitute the given value into equation 3 to have
123 = 100²/R
R = 100²/123
R = 10,000/123
R = 81.3ohms
The resistance of the computer is 81.3ohms
This problem can be solved based on the rule of energy conservation, as the energy of the photon covers both the energy needed to overcome the binding energy as well as the energy of ejection.
The rule can be written as follows:
energy of photon = binding energy + kinetic energy of ejectection
(hc) / lambda = E + 0.5 x m x v^2 where:
h is plank's constant = 6.63 x 10^-34 m^2 kg / s
c is the speed of light = 3 x 10^8 m/sec
lambda is the wavelength = 310 nm
E is the required binding energy
m is the mass of photon = 9.11 x 10^-31 kg
v is the velocity = 3.45 x 10^5 m/s
So, as you can see, all the parameters in the equation are given except for E. Substitute to get the required E as follows:
(6.63x10^-34x3x10^8)/(310x10^-9) = E + 0.5(9.11 x 10^-31)(3.45x10^5)^2
E = 6.41 x 10^-16 joule
To get the E in ev, just divide the value in joules by 1.6 x 10^-19
E = 4.009 ev
Answer:
The right answer is "1010 V/m".
Explanation:
The given values are:
Intensity,
Now,
The electric field's maximum value will be:
= 
On substituting the values in the above formula, we get
= 
= 
= 
