For an infinitely long wire bent into a semi-circular shape, we can use the Biot-Savart law to find the magnetic field at the center of the semi-circular loop.
The formula for the magnetic induction along the axis of a semi-circular loop of current is given by: \[ B = \frac{\mu_0 I}{4r} \] Where: - \( \mu_0 \) is the permeability of free space, - \( I \) is the current, - \( r \) is the radius of the semi-circular loop. Given that the radius is 1 m and the magnetic field along the axis is directed along the line passing through the center of the semi-circle, the magnitude of the magnetic induction is \( \frac{\mu_0 I}{4r} \).
Thus, the magnetic induction is \( \frac{\mu_0 I}{4r} \, \text{T} \).
The difference in energy levels of an electron at two excited levels is 13.75 eV. If it makes a transition from the higher energy level to the lower energy level then what will be the wavelength of the emitted radiation?
Given:
$ h = 6.6 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{s}^{-1} $, $ c = 3 \times 10^8 \, \text{ms}^{-1} $, $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $
200 ml of an aqueous solution contains 3.6 g of Glucose and 1.2 g of Urea maintained at a temperature equal to 27$^{\circ}$C. What is the Osmotic pressure of the solution in atmosphere units?
Given Data R = 0.082 L atm K$^{-1}$ mol$^{-1}$
Molecular Formula: Glucose = C$_6$H$_{12}$O$_6$, Urea = NH$_2$CONH$_2$