Questions
A charged particle is moving in a gravity free space in a uniform magnetic field. If an on instant of time velocity vector is and acceleration vector is then magnitude of acceleration at that instant is
detailed solution
Correct option is D
Negative force F→=QV→×B→∴ acceleration a→=F→m=QmV→×B→Hence a→ is perpendicular to V→∴a→.V→=0⇒16-4×b=0⇒b = 4 ∴a→=32+42m/s2= 5 m/s2Talk to our academic expert!
Similar Questions
A charged beam consisting of electrons and positrons enter into a region of uniform magnetic field B perpendicular to field, with a speed v. The maximum separation between an electron and a positron will be (mass of electron = mass of positron = m)
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests