# Force on Moving Charge in Magnetic Field

Let us consider a charge q moves with the velocity v through the magnetic field B as shown in figure. The magnetic force exerted by the charge is F.

Now, according to Lorentz, the force experienced by a moving charge in a uniform magnetic field is directly proportional to:

- Strength of magnetic field i.e., F ∝ B……(i)
- Magnitude of the charge i.e., F ∝ q…….(ii)
- Velocity of the charge i.e., F ∝ v……(iii)
- Angle between V and B i.e., F ∝ sinθ ……(iv)

Combining all the equations, we get:

F ∝ Bqvsinθ

F = k.Bqvsinθ Where k is a constant whose value is found to be 1.

∴ F = Bqvsinθ

Vectorically, $\vec F= q ( \vec v \times \vec B)$

Here, the direction of force is perpendicular to the plane of V and B. This force is also called as Lorentz magnetic force.

**Case 1:**

when θ =0°,

F = BqVsin0°=0

**Case2:**

when θ=90°

F = BqVsin90° =BqV

**Case3:**

When θ=180°

F = BqV180°

F = 0

**Unit of B: **The SI unit of magnetic field strength is Tesla.

We know,

$F= Bqvsin \theta$

$\text{or} B = \frac{F}{qv sin \theta } $

When charge (q) = 1 Colomb, velocity (v) = 1 metre per second and force = 1 N then,

$B = \frac{1}{1 \times 1 \times 1} = 1 \text{Tesla}$

So, the magnetic field strength at a point is 1 T if 1 charge is moving through 1 metre per second at right angle to the magnetic field and experiences 1 N magnetic force.

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