Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Answer:
Explanation:
Since the wires attract each other , the direction of current will be same in both the wires .
Let I be current in wire which is along x - axis
force of attraction per unit length between the two current carrying wire is given by
x 
where I₁ and I₂ are currents in the wires and d is distance between the two
Putting the given values
285 x 10⁻⁶ = 10⁻⁷ x 
I₂ = 16.76 A
Current in the wire along x axis is 16.76 A
To find point where magnetic field is zero due the these wires
The point will lie between the two wires as current is in the same direction.
Let at y = y , the neutral point lies
k 2 x
= k 2 x 
25.5y = 16.76 x .3 - 16.76y
42.26 y = 5.028
y = .119
= .12 m
Answer:
4 A
Explanation:
The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

where
V is the voltage
R is the resistance
I is the current
The equation can also be rewritten as

from which we see that the current is inversely proportional to the resistance, R.
In this problem, the initial current is I = 8 A. Then the resistance is doubled:
R ' = 2R
So the new current is

so the current is halved.
We can calculate the density of the balloon as follows:

Therefore, the balloon will fall
Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.