Question

A small block of mass m and charge q is pulled (from rest) by horizontal force F on a smooth horizontal ground as shown in the figure. A uniform magnetic field B, directed into the plane exists in the region. The normal force between ground and the block would become zero at time

Answer 1

Given:

Mass of the block = m

Charge of the block = q

Force acting on it = F ( x direction)

Magnetic field =  B into the plane, parallel to ground. ( z direction)

To Find:

The time at which the normal force between ground and the block would become zero.

Solution:

• Due to the motion of the charged block, the magnetic field exert a force on the charge.
• Its direction can be found out by right hand rule.
• Point the thumb along the external force.
• Point the index finger along magnetic field.
• Then point middle finger perpendicular to both the fingers.
• Its direction gives the direction of force on the charge by the magnetic field.

Therefore in this case we can observe that the direction of magnetic force will be upwards in the y direction.

• Let FF be the force due to magnetic field.
• FF =qVB  - (1),
• V = velocity of particle.

Here the mass experiences external acceleration in x direction.

• F = ma
• a = F/ m -(2)

We know if V is the velocity of particle,

• a = dV/dt
• V= at
• V = Ft/m - (3)

Magnetic force in the upward direction :

• FF = qBFt/m

For the normal force between the ground and block to become zero, the net force should be 0, in the vertical direction.

• FF = mg
• qBFt/m = mg
• t = m²g/qBF

Therefore

The time when normal force between ground and the block would become zero is m²g/qBF