# The figure (see attached) shows a long straight wire that carriers a 50-A current from left to right.  An electron, which is D = 2 μm from the wire is moving to the right at a speed of v =...

The figure (see attached) shows a long straight wire that carriers a 50-A current from left to right.

An electron, which is D = 2 μm from the wire is moving to the right at a speed of v = 5x10^6 m/s` <br> `

Determine the magnitude of the magnetic force acting on the electron at this moment.

a - 1pN

b - 2pN

c - 3pN

d - 4pN

e - 5pN

f - None

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atsuzuki | (Level 1) Adjunct Educator

Posted on

In order to answer this question, let us first list the data given:

a)    Current (I) in the wire from Left to Right: I = 50 A (amperes) - Already in the SI units.

b)   Distance D of the moving electron from wire: D = 2`mum`

. This needs to be converted into SI units before we do the calculations, `D=2*10^(-6)m`

` <br data-mce-bogus="1"> `

c)    Speed of the electron (to the right): `v=5*10^6 (m/s)`

` `  .

d)   Not given, but needed, permeability of vacuum: μ0 = 4 x 10-7 N/A2.

e)    Not given, but needed, electron charge: e = 1.6 x 10-19 C.

Question asked: magnitude of the magnetic force on the electron.

To solve this problem, we first have to know the magnitude of the  magnetic field at the point where the electron is. This is given by the following formula:

`B=(mu_0*I)/(2*pi*B)`

.

Plugging in the values we have:

`B=(4*pi*10^(-7)(N/(A^2))*50(A))/(2*pi*2*10^(-6)(m))=5(N/(A*m))=5((N*s)/(C*m))`

.

According to the right hand rule, this magnetic field where the electron is positioned points in the direction perpendicular to the plane of the paper, pointing away from the paper. The unit N/A.m is also known as tesla (T).

Now we need to evaluate the force acting on this moving electron. This is given by

`vecF_(mag)=evec v ^^ vec B`

` <br data-mce-bogus="1"> `

According to the vector product convention, this force is perpendicular to the direction of the electron speed and the direction of the magnetic field. The rule gives the direction of this force as perpendicular to the wire, pointing towards the wire. However, since the electron has a negative charge, the direction of the resulting force is reversed, i.e, points away from the wire, perpendicularly to the wire. However, since the problem asks for only the magnitude of the magnetic force, we can set

`F_(mag)=evBsin90^0=evB`

` <br data-mce-bogus="1"> `

` `

Now it is a question of plugging into this the values we have:

`F_(mag)=1.6*10^(-19)(C)*5*10^6 (m/s)*5((N*s)/(C*m))=40*10^(-13) (N).`

``

In the above, we have used the fact that the current of 1 A = 1 C/s. in the unit of the magnitude of the magnetic field. Since 1 pN = 10-12 N, the answer is that the magnitude of the magnetic field is

`F_(mag)=4 pN.`

So the correct answer to the problem is alternative d).

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The magnitude of a magnetic field around a wire carrying current I at a radial distance r is given by `B = (4*pi*10^-7*I)/(2*pi*r) = (2*10^-7*I)/r` . If the wire is held in the right hand with the thumb pointing in the direction of flow of current, the fingers trace the direction of the magnetic field.

The magnetic field around the wire at a distance of 2*10^-6 m is `(2*10^-7*50)/(2*10^-6)` = 5 T

The force exerted by a magnetic field B on a particle with charge q moving at speed v is equal to `F = q*v xx B`

Here, the magnetic field is perpendicular to the direction of the particle's velocity

The force exerted by this field on the electron moving at v = 5*10^6 m/s is equal to `5*10^6*5*1.60217657*10^-19 ~~ 40` pN

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