A velocity selector consists of electric field $$\vec{E}=E \,\hat{k}$$ and magnetic field $$\vec{B}=B \,\hat{j}$$ with $$B=12 \,m T$$. The value of $$E$$ required for an electron of energy $$728 \,\mathrm{e} V$$ moving along the positive $$x$$-axis to pass undeflected is :

(Given, mass of electron $$=9.1 \times 10^{-31} \mathrm{~kg}$$ )

The magnetic field of a plane electromagnetic wave is given by :

$$ \overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}$$.

The amplitude of the electric field would be :

Light wave travelling in air along x-direction is given by $${E_y} = 540\sin \pi \times {10^4}(x - ct)\,V{m^{ - 1}}$$. Then, the peak value of magnetic field of wave will be (Given c = 3 $$\times$$ 10⁸ ms^$$-$$1)

The rms value of conduction current in a parallel plate capacitor is $$6.9 \,\mu \mathrm{A}$$. The capacity of this capacitor, if it is connected to $$230 \mathrm{~V}$$ ac supply with an angular frequency of $$600 \,\mathrm{rad} / \mathrm{s}$$, will be :