Apositive charge, q1, of 5 µc is 3 × 10–2 m west of a positive charge, q2, of 2 µc. what is the magnitude and direction of the electrical force, fe, applied by q1 on q2? a) magnitude: 3 n direction: east b) magnitude: 3 n direction: west c) magnitude: 100 n direction: east d) magnitude: 100 n direction: west

The magnitude and direction of the electrical force, Fe, applied by q1 on q2 is:

C. magnitude: 100 N

direction: east

The correct answer option is C) magnitude: 100 N

, direction: east.

We know the formula of electrical force:

F = 1/4πε * q₁q₂ / r²

and we know the following values:

q₁ = 5 x 10⁻⁶ C

q₂ = 2 x 10⁻⁶ C

r = 3 x 10⁻² m west

So we can substitute these values in the formula to find the magnitude of F:

F = (9 x 10⁹) (5 x 10⁻⁶) (2 x 10⁻⁶)  / (3 x 10⁻²)²

F = 100 N  (East of positive charge)

Therefore, the magnitude of F is 100N applied by q1 and q2, which directs towards East of positive charge.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: