Projectile Motion Of Charged Particle 2

At the moment the object (of charge +q and mass m) enters the magnetic field, there could be three forces acting on it. Since the object is in uniform motion in the magnetic field, two of these forces must cancel each other and the third should equal zero.

First, the force due to gravity will act on the mass m and will be directed DOWNWARDS (F = mg).

Second, there will be a magnetic force acting on it due to it's velocity in the horizontal direction (Vo COS(45) because the velocity in the horizontal direction will not have changed ...if this confuses you, learn about projectiles.). Using the Right Hand Rule, we can see that this magnetic force will be directed UPWARDS (F = q Vo COS(45) B because magnetic force = qvB).

Thirdly, there could be magnetic force due to the vertical component of it's velocity. We know using the right hand rule that this force acts in LEFTWARDS. But, there is no other force that could act in the rightward direction and cancel this force. So the magnetic force in the leftward direction should be zero and this is possible because the vertical component of velocity is decreasing constantly from the moment the projectile is thrown due to gravity.)

Now, the UPWARDS force must equal the DOWNWARDS force to enable the body to be in uniform motion.

mg = q Vo cos(45) B

Solving for B, we get the answer.

Force acting on charge particle due to motion in a uniform magnetic field (in up words) is equal to the force (down in direction) acting on the charged particle of mass m due to earth's gravitational field i.e., q(VXB) = mg , where V is the velocity of charged particle in Uniform Magnetic Field(B)

Force acting on the body inside the magnetic field is mg in y direction.Velocity of the body in y direction is $v_0/\sqrt{2}$ .

$mg=q \times(v_0/\sqrt{2}) \times B$ where B is the intensity of the magnetic field.

I think the ques is wrong. The charge on the particle must be negative. Then only the forces along y direction balance each other