एक कण पर गुरुत्वीय बल mg के अतिरिक्त एक अन्य बल vec(F)=vec(v)xx vec(A) लग रहा है, जहाँ vec(v) कण...

What is the angle between the v ectprs (vec A xx vec B) and (vec Bxx vec A) ?

Foe an arbitrary motion in sparce, which ot the following relations are true : (a0 vec (average ) = 1/2 [ vec v (t_10 + vec v (t_2)] (b) vec _(average ) = [vec r 9t_2) - vec r (t-1 0 ] // (t_2 -t_10 (c ) vec v(t) = vec v (0) + vec a t (d) vec r 9t) = vec r90) + vec v (0) t + ( 1//2 ) vec a t^2 (e) vec a_(average) =[ vec v( t_2) - vec v (t-2 ) ] // (t_2 -t_1)1 The average stands for average of the wuantity over the time interval t_1 and t_2 ].

Magnetic force on a charged particle is given by vec F_(m) = q(vec(v) xx vec(B)) and electrostatic force vec F_(e) = q vec (E) . A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by vec(E) = (10 hat(i)) N//C for x = 1.8 m and vec(B) = -(5 hat(k)) T for 1.8 m le x le 2.4 m A screen is placed parallel to y-z plane at x = 3 m . Neglect gravity forces. The speed with which the particle will collide the screen is

Magnetic force on a charged particle is given by vec F_(m) = q(vec(v) xx vec(B)) and electrostatic force vec F_(e) = q vec (E) . A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by vec(E) = (10 hat(i)) N//C for x = 1.8 m and vec(B) = -(5 hat(k)) T for 1.8 m le x le 2.4 m A screen is placed parallel to y-z plane at x = 3 m . Neglect gravity forces. Time after which the particle will collide the screen is (in seconds)

Magnetic force on a charged particle is given by vec F_(m) = q(vec(v) xx vec(B)) and electrostatic force vec F_(e) = q vec (E) . A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by vec(E) = (10 hat(i)) N//C for x = 1.8 m and vec(B) = -(5 hat(k)) T for 1.8 m le x le 2.4 m A screen is placed parallel to y-z plane at x = 3 m . Neglect gravity forces. y-coordinate of particle where it collides with screen (in meters) is

As a charged particle 'q' moving with a velocity vec(v) enters a uniform magnetic field vec(B) , it experience a force vec(F) = q(vec(v) xx vec(B)). For theta = 0^(@) or 180^(@), theta being the angle between vec(v) and vec(B) , force experienced is zero and the particle passes undeflected. For theta = 90^(@) , the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal force (mv^(2)//r) . For other values of theta (theta !=0^(@), 180^(@), 90^(@)) , the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle that carries a charge of magnitude q and has a mass 4 xx 10^(-15) kg is moving in a region containing a uniform magnetic field vec(B) = -0.4 hat(k) T . At some instant, velocity of the particle is vec(v) = (8 hat(i) - 6 hat(j) 4 hat(k)) xx 10^(6) m s^(-1) and force acting on it has a magnitude 1.6 N If the coordinates of the particle at t = 0 are (2 m, 1 m, 0), coordinates at a time t = 3 T, where T is the time period of circular component of motion. will be (take pi = 3.14 )