Question

[
F=frac{1}{A_{0} cos theta}=frac{2 A}{1 mathrm{cm}^{2} cdot cos (60)}=frac{2 times 2}{10^{-11} times 1}
]
( I=frac{I}{A cdot cos (90-theta)}=frac{2 A}{10 mathrm{m}^{2} cos (30)}=frac{2 times 2}{10^{-4} times sqrt{3}} frac{A}{pi}=frac{A}{M^{2}} )
( F=frac{4}{sqrt{3} times 10^{4}} A / m^{2} )

# 4. (C) increases (D) varies randomly The cross-sectional area of the plane shown in the figure is equal to 1 cm. 2A current flows through a conductor. The current density at point P in the conductor will be ...... Conductor, 1=2A A : 10Am B E109Am oy - 10Am (D) * - 10-Am ? ...a rhit of radius 53 Y 111 m with a constant

Solution