Two particles `A and B` of masses ` m_(A) and m_(B)` respectively and having the same charge are moving in a plane . The speeds of the particles are ` v_(A) and v_(B)` respectively and the trajectories are as shown in the figure. Then
A
` m_(A) v_(A) lt m_(B) v_(B) `
B
` m_(A) v_(A) gt m_(B) v_(B)`
C
` m_(A) v_(A) lt m_(B) and v_(A) lt v_(B) `
D
` m_(A) = m_(B) and v_(a) = v_(B)`
Text Solution
Verified by Experts
The correct Answer is:
B
KEY COCNCEPT: When a charged particle is moving at right angles to the magnetic field then a force acts on it which behaves as a centripetal force and moves the particle in circular motion . :. ` (m_(A)v^(2)A)/(r_(A)) = q .v_(A)B :. ( m_(A)v_(A))/(r_(A) = q_(B) ` ` (m_(B)v_(B))/(r_(B) = q^(B)` rArr `(m_(A)v_(A))/(r_(A)) = ( m_(B)v_(B))/(r_(B))` ` r_(A)gt r_(B)` rArr m_(A)v_(A) gt m_(B) v_(B) `
Topper's Solved these Questions
MOVING CHARGES AND MAGNETISM
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise MCQs(d )|1 Videos
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise JEE Main And Advanced|209 Videos
Similar Questions
Explore conceptually related problems
Two particle A and B of masses m_(A) and m_(B) respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are v_(A) and v_(B) respectively and the trajectories are as shown in the figure. Then-
Two particles A and B of masses m_(A)andm_(B) respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are v_(A)andv_(B) respectively and the trajectories are as shown in the figure. Then -
Two particles A and B of masses m_A and m_B respectively and having the same charge are moving ina plane. A uniform magnetic field exists perependicular to this plane. The speeds of the particles are v_A and v_B respectively and the trajectories are as shown in the figure. Then,
Two particles A & B of mass m_(1) and m_(2) respectively start moving from O with speeds v_(1) , and v_(2) . A moves towards the plane mirror and B moves parallel to mirror horizontally. The mirror is in y-z plane. The absolute-speed of image of centre of mass of the system (image of A + image of B)is:
Particle A makes a perfectly elastic collision with anther particle B at rest. They fly apart in opposite direction with equal speeds. If the masses are m_(A)&m_(B) respectively, then
Two particles A and B with particle A having mass m, are moving towards each other under their mutual force of attraction. If their speeds at certain instant are v and 2v respectively, then find the K.E. of the system.
The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit. There is one particle A having charge q and mass m. There is another particle B of charge eight times and mass two times to that of A. Both the particles are accelerated through the same potential difference. Let speeds acquired by the particles A and B be v_(A) and v_(B) respectively, then calculate v_(B)//v_(A) .
SUNIL BATRA (41 YEARS IITJEE PHYSICS)-MOVING CHARGES AND MAGNETISM-MCQs(d )
