A charged particle moves with velocity →v=a^i+d^j in a magnetic field →B=A^i+D^j. The force acting on the particle has magnitude F. Then,

F=0, if aD=dA

F=0, if aD=−dA

F∝(a2+b2)1/2×(A2+D2)1/2

F=0, if aA=−dD

A

F∝(a2+b2)1/2×(A2+D2)1/2

B

F=0, if aA=−dD

C

F=0, if aD=dA

D

F=0, if aD=−dA

Open in App

Solution

Verified by Toppr

→F=q(→v×→B)

Was this answer helpful?

3

Similar Questions

Q1

A charged particle moves with velocity →v=a^i+d^j in a magnetic field →B=A^i+D^j. The force acting on the particle has magnitude F. Then,

View Solution

Q2

A charged particle with velocity →v=x^i+y^j moves in a magnetic field →B=y^i+x^j. The magnitude of magnetic force acting on the particles F. Among the following the correct statement(s) is/are :

View Solution

Q3

A Positive charge particle with velocity →v=x^i+y^j moves in a magnetic field →B=y^i+x^j. The magnitude of magnetic force acting on the particles is F. Which one of the following statements are correct :

View Solution

Q4

A charged particle with velocity →v=x^i+y^j moves in a magnetic field →B=y^i+x^j. The force acting on the particle has magnitude F. Which one of the following statements is/are correct?

View Solution

Q5

If the expression in vector form, for the magnetic force →F acting on a charged particle moving with velocity →V in the presence of a magnetic field →B is given by →F=a∗q→v×→r. Find a.