Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
The correct answer would be left
Answer:
he spring provides the controlling torque. The air friction induces the damping torque, which opposes the movement of the coil. The repulsion type instrument is a non-polarized instrument, i.e., free from the direction of current passes through it. Thus, it is used for both AC and DC
Answer:
d = 1.047 mm
Explanation:
given,
diameter of the wire = 2.0-cm
length of solenoid = 15 cm = 0.15
Current in the wire = I = 2.5 A
magnetic field = B = 3.0 mT
Magnetic field inside the solenoid


N x d = l




d = 1.047 x 10⁻³ m
d = 1.047 mm
diameter of the wire is d = 1.047 mm