Question

A proton, traveling with a velocity of 4.5 x 10-5 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.0 x 10-14 N and a direction of due south. What are the magnitude and direction of the magnetic field causing the force? Repeat the aforementioned questions assuming the proton is replaced by an electron.

Answer 1

The magnitude of the field is 1.112 T

The direction in case of proton is
`\hat{k}`, i.e., outwards

The direction in case of proton is
`\hat{- k}`, i.e., inwards

Solution:

As per the question:

Velocity of the proton,
`\vec{v} = 4.5* 10^(5)\hat{i}\ m/s`

Magnetic force,
`\vec{F}8.0* 10^(- 14)\hat{- j}\ N`

Charge on proton, q =
`1.6* 10^(- 19)\ C`

Charge on electron, e =
`- 1.6* 10^(- 19)\ C`

Now,

The magnitude of the mgnetic field can be given by the formula of Lorentz force:

`\vec{F} = q(\vec{v}* \vec{B})`

For maximum force, the magnitude of the force is given by:

F = qvB

`B = (F)/(qv)`

`B = (8.0* 10^(- 14))/(1.6* 10^(- 19)* 4.5* 10^(5)) = 1.112* T`

Now, for the direction of the field:

`\hat{- j} = \hat{i}* \vec{B})`

The direction that we get from the above eqn is
`\hat{k}`

Now, in case of electron:

`B = (F)/(qv)`

`B = (8.0* 10^(- 14))/(1.6* 10^(- 19)* 4.5* 10^(5)) = 1.112* T`

Now, for the direction of the field:

`\hat{- j} = -\hat{i}* \vec{B})`

The direction that we get from the above eqn is
`\hat{- k}`, i.e., directed inwards.