A beam of charged particles of charge q and a mass m are accelerated from rest through a potential difference of 100V. They pass through crossed electric and magnetic fileds which together produce null deflection .If these electronic and magnetic fields are respectively `15xx10^(3)Vm^(-1)and 5Wbm^(-2)`,then q/m has a value, in`C kg^(-1)`,equal to

Options

A. `9xx10^(4)`
B. 15
C. `4.5xx10^(4)`
D. `4.5xx10^(4)`

Answer

C

Explanation

Potential difference ,V=100V
Electronic field ,`E=15xx10^(3)Vm^(-1)`
Magnetic field ,`B=5Wbm ^(-2)`
The expression for K.E of a particle is given as, K.E=`(1)/(2)mv^(2)`(because Kinetic energy =qv)
`impliesqV=(1)/(2)mv^(2)`
`implies (q)/(m)=(1)/(2)(v^(2))/(V)`
Where ,
V-Velocity of charged particle=`(K)/(B)`
=`(15xx10^(3))/(5)`
`V=3xx10^(3)ms-(1)`
`implies (q)/(m)=(1)/(2)xx(3xx10^(3))^(2)/(100)`
`(1)/(2)xx(9xx10^(6))/100`
`(q)/(m)=4.5xx10^(4)ckg^(-1)`