Question

A particle with a charge of 11.2 C moves with a velocity of 11.7 m/s toward geographic West (Be = 5.01 × 10-S T). If the Lorentz force on the particle is 9.12 × 10-9 N, what is the magnitude and direction of the electric field?

Answer 1

To find the magnitude and direction of the electric field, we can use the equation for the Lorentz force
`\displaystyle\sf (F_(Lorentz))` experienced by a charged particle moving in a magnetic field:

`\displaystyle\sf F_(Lorentz)=q\cdot v\cdot B`,

where
`\displaystyle\sf F_(Lorentz)` is the Lorentz force,
`\displaystyle\sf q` is the charge of the particle,
`\displaystyle\sf v` is the velocity of the particle, and
`\displaystyle\sf B` is the magnetic field.

In this case, the Lorentz force
`\displaystyle\sf F_(Lorentz)` is given as
`\displaystyle\sf 9.12* 10^(-9)\,N`, the charge
`\displaystyle\sf q` is
`\displaystyle\sf 11.2\,C`, and the velocity
`\displaystyle\sf v` is
`\displaystyle\sf 11.7\,m/s`.

We can rearrange the equation to solve for the magnitude of the magnetic field
`\displaystyle\sf B`:

`\displaystyle\sf B=(F_(Lorentz))/(q\cdot v)`.

Substituting the given values, we have
`\displaystyle\sf B=(9.12* 10^(-9)\,N)/(11.2\,C\cdot 11.7\,m/s)`.

Calculating this, we find
`\displaystyle\sf B\approx 6.86* 10^(-9)\,T`.

Therefore, the magnitude of the magnetic field is
`\displaystyle\sf 6.86* 10^(-9)\,T`.

To determine the direction of the electric field, we need to consider the Lorentz force equation:

`\displaystyle\sf F_(Lorentz)=q\cdot v\cdot B`.

The Lorentz force is perpendicular to both the velocity
`\displaystyle\sf v` and the magnetic field
`\displaystyle\sf B`. Given that the particle is moving toward the geographic West, the magnetic field
`\displaystyle\sf B` must be directed downward (toward the Earth's surface) to exert a force towards the East (perpendicular to the velocity).

Therefore, the direction of the electric field is downward.

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