# Find the equation of this magnetic field

• Istiak
In summary, the question asks you to find the magnetic field of an electron that enters a square region ABCD with area a. You are asked to find the minimum magnetic field strength that will cause the electron to exit the square in the direction opposite from the direction it was fired in.f

#### Istiak

Homework Statement
An electron with velocity v =##2 \times 10^6 \\ \mathrm {ms^-1}##
enters a square region ABCD (of area a = ##1 \\ \mathrm cm^2##
along one of its side AB. Inside the region there is a magnetic field B perpendicular to the
area of the square. Find the minimum value of magnetic field for which the electron will
come out of the square with a velocity parallel to its initial velocity(parallel doesn't
necessarily mean in the same direction).
Relevant Equations
##\vec B = \frac{\mu_0 q \vec v\times \hat r}{4\pi r^2}##
When I try following numbers from internet then I don't get an expected answer.

## \mu_0 = 1.25663706 × 10-6 m kg s^{-2} A^{-2}##
##q =1.60217662 × 10^{-19} coulombs ##
##r=2.82x10^{-15} m##
Velocity of that electron is given in question

##\vec v= 2 \times 10^6 \\ \mathrm{ms^{-1}}##

Since magnetic field is perpendicular to the surface that's why I took ##\vec v \times \hat r=||v||## Wait a minute, Magnetic field is perpendicular but not velocity and velocity is parallel so ##\vec v \times \hat r=0##. But if I tried it then I would get "nothing". Did I take wrong equation? Or there's some concept which I haven't figured out?

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Where did you get the value of ##r## from?

Edit: Also, why do you think the field is given by that formula?

Where did you get the value of ##r## from?
That appears to be the radius of an electron!

Where did you get the value of ##r## from?

Edit: Also, why do you think the field is given by that formula?
I found the ##r## from internet where it is, radius of an electron. I found that equation in a book so I believe magnetic field is given by that equation. But I worry I can't derive one for a square.

##\oint \vec B \cdot d\vec l = \mu_0 I_{enc}##
##\vec B =\frac{\mu_0 I_{enc}}{2r}##

My given equation was from (steady current law)
##\vec B = \frac{\mu_0 }{4\pi}\int\frac{\vec K \times \hat r}{r^2}da\prime##
##=\text{given at top of the thread.}##

I found the ##r## from internet where it is, radius of an electron. I found that equation in a book so I believe magnetic field is given by that equation. But I worry I can't derive one for a square.

##\oint \vec B \cdot d\vec l = \mu_0 I_{enc}##
##\vec B =\frac{\mu_0 I_{enc}}{2r}##

My given equation was from (steady current law)
##\vec B = \frac{\mu_0 }{4\pi}\int\frac{\vec K \times \hat r}{r^2}da\prime##
##=\text{given at top of the thread.}##
Does the question ask you to find the magnetic field of an electron?

Does the question ask you to find the magnetic field of an electron?
An electron with velocity v =##2 \times 10^6 \\ \mathrm {ms^-1}##
enters a square region ABCD (of area a = ##1 \\ \mathrm cm^2##
along one of its side AB. Inside the region there is a magnetic field B perpendicular to the
area of the square. **Find the minimum value of magnetic field** for which the electron will
come out of the square with a velocity parallel to its initial velocity(parallel doesn't
necessarily mean in the same direction).

Take a look at what I bold...!

Find the minimum value of magnetic field for which the electron will
come out of the square with a velocity parallel to its initial velocity(parallel doesn't
necessarily mean in the same direction).
IMO, it would be better if the question stated simply what it wants you to find. You are aked to calculate the minimum strength of the magnetic field so that the electron exits the square in the direction opposite from the direction it was fired in.

To do the problem you must know or calculate the shape of the trajectory of a charged particle in a uniform magnetic field.