A particle of mass 0.6 ,g and having charge o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: Mass $m = 0.6 \,g = 0.6 	imes 10^{-3} \,kg$, Charge $q = 25 \,nC = 25 	imes 10^{-9} \,C$, Velocity $v = 1.2 	imes 10^4 \,ms^{-1}$, Accel

Vedclass · Accepted Answer

$20 \,T$

Vedclass · Answer

(C) Given: Mass $m = 0.6 \,g = 0.6 	imes 10^{-3} \,kg$, Charge $q = 25 \,nC = 25 	imes 10^{-9} \,C$, Velocity $v = 1.2 	imes 10^4 \,ms^{-1}$, Acceleration due to gravity $g = 10 \,ms^{-2}$.Since the particle moves with a uniform velocity, the net force acting on it must be zero.This implies that the magnetic force $F_m$ must balance the gravitational force $F_g$ acting on the particle.$F_m = F_g$$Bqv = mg$$B = \frac{mg}{qv}$Substituting the values:$B = \frac{0.6 	imes 10^{-3} 	imes 10}{25 	imes 10^{-9} 	imes 1.2 	imes 10^4}$$B = \frac{6 	imes 10^{-3}}{30 	imes 10^{-5}}$$B = \frac{6 	imes 10^2}{30} = \frac{600}{30} = 20 \,T$Therefore, the magnetic induction is $20 \,T$.

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

Similar Questions

$A$ particle of charge $e$ and mass $m$ moves with a velocity $v$ in a magnetic field $B$ applied perpendicular to the motion of the particle. The radius $r$ of its path in the field is

If a proton, deuteron, and $\alpha$-particle are accelerated by the same potential difference and enter perpendicular to a magnetic field, then the ratio of their kinetic energies is:

$A$ proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

$A$ proton moving with a velocity of $2.5 \times 10^7 \, m/s$ enters a magnetic field of intensity $2.5 \, T$ making an angle of $30^o$ with the magnetic field. The force on the proton is:

$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

Vedclass Test Series

Exam Paper Generator

Online Exam Module