An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :
(1) h eB
Magnetic force gives the centripetal force: evB= mv2r mv=eBr
[figure]
Bohr: mvr= nh2 eBr2= nh2 r= nh2 eB
First excited state: n=2 r= h eB
Answer: (1)
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2 105 m s-1. When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is x 104 N/C. The value of x is . Take the mass of the proton =1.6 10-27 kg.
Two long parallel wires X and Y, separated by a distance of 6 cm , carry currents of 5 A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is x 10-5 T. The value of x is . Take permeability of free space as 0=4 10-7 SI units. [figure]