Awire of a certain length is carrying a current of 2.0 a. it is placed at an angle of 60° with respect to a magnetic field of strength 0.20 t. if the wire experiences a force of 0.40 n, what is the length of the wire?

1.2 m

Explanation:

i = current flowing through the wire = 2.0 A

θ = Angle between the length of wire and magnetic field = 60° deg

B = magnitude of magnetic field = 0.20 T

L = Length of the wire = ?

F = Magnitude of magnetic force experienced by the wire

Magnetic force on the wire is given as

F = i B L Sinθ

0.40 = (2.0) (0.20) L Sin60

L = 1.2 m

at right angles to each other

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j