A particle of mass ‘m’ and charge ‘q’ is projected with a velocity v0 in a viscous medium, where a uniform and constant magnetic field of induction B exists everywhere in a direction perpendicular to the direction of projection of the particle. The force of viscous drag on the particle is given by the law f→=−bv→, where ‘b’ is a positive constant and v→ is velocity of the particle. The distance travelled by the particle during a time interval from the instant of projection until velocity vector turns by 2π radians is mv0b(1−enπbqB). Find ‘n’?
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 2.00.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Along the tangent net force is F=−bv⇒mdvdt=−bvDue to magnetic force, the particle move in circular path of radius R.∴v=dldt=Rdθdt ⇒mdv=−bRdθ⇒dv=−bmmvqBdθ⇒ V0Vdvv=−bqB 02πdθ⇒v=v0e−b2πqBNow, againF=−bv⇒mvdvds=−bv⇒m∫v0vdv=−b∫0Sds⇒mv−v0=−bSSo, distance covered S=mbv0−v=mv0b1−e2πbqB