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## How to Calculate the Magnetic Force Perpendicular to the Direction of a Charge

**Step 1:** Read the problem and identify the values of the charge *q*, the value of the velocity of the particle *v* and the value of the magnetic field *B*.

**Step 2:** Put these values into the equation: {eq}\rm{F=|q|vB}
{/eq}.

Note: *|q| *is the absolute value of the charge *q*.

**Step 3:** Using this equation calculate the force *F*.

**Step 4:** Check your units!

## What is the magnetic force?

Just like a charge that does not move produces an electric field, a moving charge produces a magnetic field. An object inside a magnetic field experiences a force that we call magnetic force.

**Force:** A push or a pull that can change an object's motion. The standard unit for force is a Newton (N).

**Charge:** The property of an object that allows it to be affected inside an electric or magnetic field. The standard unit of charge (also referred to as electric charge) is 1 Coulomb.

**Velocity:** The distance that an object travels in a unit of time. The velocity of an object includes the direction that the object travels to. The standard unit of velocity is m/s.

**Magnetic field:** An area in space where magnetic forces exist. We can visualize a magnetic field when, for example, we place iron fillings on a piece of paper and a magnet underneath the paper. The standard unit of the magnetic field is 1 Tesla. {eq}\rm{\left(T=\frac{N \, s}{C \, m }\right)}
{/eq}

We will now solve two problems (step-by-step) to enforce our understanding as to how to calculate the magnetic force perpendicular to the direction of a charge.

## Examples of How to Calculate the Magnetic Force Perpendicular to the Direction of a Charge

**Example 1**

A proton is moving perpendicularly at {eq}\rm{4.5 \times 10^{6} m/s} {/eq} through a magnetic field of 1.5 T. What is the magnetic force that the proton experiences? Charge of a proton: {eq}\rm{1.602 \times 10^{-19} C} {/eq}

**Step 1:** Read the problem and identify the values of the charge *q*, the value of the velocity of the particle *v* and the value of the magnetic field *B*.

Charge: {eq}\rm{1.602 \times 10^{-19} C} {/eq}

Velocity: {eq}\rm{4.5 \times 10^{6} m/s} {/eq}

Magnetic field: Magnetic field: 1.5 T

**Step 2:** Put these values into the equation: {eq}\rm{F=|q|vB}
{/eq}.

{eq}\rm{F=|1.602 \times 10^{-19} C|(4.5 \times 10^{6} m/s)(1.5 T)} {/eq}

**Step 3:** Using this equation calculate the force *F*.

{eq}\rm{F=|1.602 \times 10^{-19} C|(4.5 \times 10^{6} m/s)(1.5 T)} {/eq}

Since {eq}\rm{T=\frac{N \, s}{C \, m }} {/eq} all the units cancel except the Newtons:

{eq}\rm{F=1.1 \times 10^{-12} N} {/eq}

**Step 4:** Check your units!

Our units are correct.

Our final answer is:

{eq}\rm{F=1.1 \times 10^{-12} N} {/eq}

**Example 2**

A charged particle carrying a charge of -2.5 C is moving perpendicularly to the magnetic field of the earth with a speed of 300,000 m/s. What is the force that the particle experiences if the magnetic field of the earth is 0.00003 T?

**Step 1:** Read the problem and identify the values of the charge *q*, the value of the velocity of the particle *v* and the value of the magnetic field *B*.

Charge: -2.5 C

Velocity: 300,000 m/s

Magnetic field: Magnetic field: 0.00003 T

**Step 2:** Put these values into the equation: {eq}\rm{F=|q|vB}
{/eq}.

{eq}\rm{F=|-2.5 C| (300,000 m/s)(0.00003 T)} {/eq}

**Step 3:** Using this equation calculate the force *F*.

{eq}\rm{F=|-2.5 C| (300,000 m/s)(0.00003 T)} {/eq}

Since {eq}\rm{T=\frac{N \, s}{C \, m }} {/eq} all the units cancel except the Newtons:

{eq}\rm{F= 20 N} {/eq}

**Step 4:** Check your units!

Our units are correct.

Our final answer is:

{eq}\rm{F= 20 N} {/eq}