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

Answer:

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, $\vec{v} = 4.5\times 10^{5}\hat{i}\ m/s$

Magnetic force, $\vec{F}8.0\times 10^{- 14}\hat{- j}\ N$

Charge on proton, q = $1.6\times 10^{- 19}\ C$

Charge on electron, e = $- 1.6\times 10^{- 19}\ C$

Now,

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

$\vec{F} = q(\vec{v}\times \vec{B})$

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

F = qvB

$B = \frac{F}{qv}$

$B = \frac{8.0\times 10^{- 14}}{1.6\times 10^{- 19}\times 4.5\times 10^{5}} = 1.112\times T$

Now, for the direction of the field:

$\hat{- j} = \hat{i}\times \vec{B})$

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

Now, in case of electron:

$B = \frac{F}{qv}$

$B = \frac{8.0\times 10^{- 14}}{1.6\times 10^{- 19}\times 4.5\times 10^{5}} = 1.112\times T$

Now, for the direction of the field:

$\hat{- j} = -\hat{i}\times \vec{B})$

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