A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0\times 10^{-5}~\text{T}4.0×10 −5 T that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 cm above the wire?

Explanation:

vertical magnetic field B_v = 4 x 10⁻⁵ T.

Magnetic field due to horizontal current at point 20 cm above

= (μ₀/4π ) x (2i / R)

= 10⁻⁷ x 2 x 20/ 20 x 10⁻²

= 2 x 10⁻⁵ T

It will act  coming out of paper. Hence it will be normal to magnetic field given .

So resultant magnetic field

= √ (4² + 2²) x 10⁻⁵ T

= 4.47 X 10⁻⁵ T .

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