Question

One coulomb of point charge moving with a uniform velocity 10 \(\widehat x\) m/s enters the region x ≥ 0 having a magnetic flux density\(\overrightarrow B = \left( {10y\widehat x + 10x\widehat y + 10\widehat z} \right)T\). The magnitude of force on the charge at x = 0⁺ is ______ N.

(\({\widehat x}\), \({\widehat y}\) and \({\widehat z}\) are unit vectors along x-axis, y-axis, and z-axis, respectively.)

Answer 1

Concept:

The force acting on a charged particle moving under the presence of a magnetic field is given by:

F̅mag = q (v̅ × B̅)

v̅ = Velocity vector of the particle

B̅ = Magnetic field vector

q = Charge of the particle

Application:

Given, q = 1 C

v̅ = 10 \(\widehat x\) m/s

\(\overrightarrow B = \left( {10y\widehat x + 10x\widehat y + 10\widehat z} \right)T\)

F̅mag = q (v̅ × B̅)

F̅mag = \(1(10\;\widehat x \times \left( {10y\widehat x + 10x\widehat y + 10\widehat z} \right))\;N\)

F̅_mag = \(100x\widehat z \;+100(-\widehat y) \; N\)

F̅mag (x = 0+) = - 100 y̅ N

Magnitude of F̅_mag = \(|\sqrt{(-100)^2}| = 100\;N \)

The Lorentz force equation describes the magnitude of the force that a moving electric charge would feel as a result of being in the presence of a magnetic field B̅ and an electric field E̅.

The Force due to the presence of an electric field is given by:

F̅elec. = q × E̅ 

q = Charge of the particle

E̅ = Electric field vector

Also, the force acting on a charged particle moving under the presence of a magnetic field is given by:

F̅mag. = q (v̅ × B̅)

v̅ = Velocity vector of the particle

B̅ = Magnetic field vector

When the charge is moving in presence of both an electric and magnetic field, the force acting on it will be the sum of the two forces, i.e.

Fnet = F̅elec. + F̅mag. 

Fnet = q × E̅ + q (v̅ × B̅)

Fnet = q (E̅ + v̅ × B̅)