Answer:
Induced current, I = 18.88 A
Explanation:
It is given that,
Number of turns, N = 78
Radius of the circular coil, r = 34 cm = 0.34 m
Magnetic field changes from 2.4 T to 0.4 T in 2 s.
Resistance of the coil, R = 1.5 ohms
We need to find the magnitude of the induced current in the coil. The induced emf is given by :
Where
is the rate of change of magnetic flux,
And 
Using Ohm's law, 
Induced current, 
I = 18.88 A
So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m
The answer would be A, because the wind cannot complain, therefore it has been given a human quality.
Answer:
Explanation:
Given
velocity of driver 
=25 m/s w.r.t ground towards north
driver observes that rain is making an angle of 
with vertical
While returning 
=25 m/s w.r.t. ground towards south
suppose 
=velocity of rain drop relative car while car is going towards north
=velocity of rain drop relative car while car is going towards south
z=vector sum of 
Now from graph
therefore magnitude of z is given by
Thus rain drops make an angle of w.r.t to ground
