Question

A straight conductor carries a current of 5 A An electron travelling with a \[{{10}^{6}}m{{s}^{-1}}\] parallel to the wire at a distance of 0.1 m from the conductor, experience a force of
A. $8\times {{10}^{-20}}N$
B. $3.3\times {{10}^{-19}}N$
C. $8\times {{10}^{-18}}N$
D. $1.6\times {{10}^{-19}}N$

Answer 1

Hint: We need to know that the force on a charged particle is because of the magnetic field produced around it. So, we need to first find the magnetic field around an electron and then the force on the electron can be calculated.

Formula used:
\[B=\dfrac{{{\mu }_{o}}}{4\pi }.\dfrac{2I}{a}\]
where \[\dfrac{{{\mu }_{o}}}{4\pi }\] is a constant which has a value of ${{10}^{-7}}$, I is the current flowing through the wire and a is the distance between the wire and the electron.
$F=qvB$
Here q is the charge on a particle (here electron), v is the velocity with which it is travelling and B is the magnetic field.

Complete step by step solution:
We have been asked to find out the force acting on the electron. To proceed here, first we need to find the magnetic field around the charged particle because it’s not given in the question. So, we know that moving current produces a magnetic field and in this case, the conductor carrying 5A current will produce a magnetic field around the electron. So, for magnetic field, the formula is given by:
\[B=\dfrac{{{\mu }_{o}}}{4\pi }.\dfrac{2I}{a}\],
According to the question $a=0.1m$and I=5A. We know that \[\dfrac{{{\mu }_{o}}}{4\pi }\] is a constant which has a value of ${{10}^{-7}}$. Putting all the values in the formula we get:
\[B=\dfrac{{{\mu }_{o}}}{4\pi }.\dfrac{2(5)}{0.1} \\ \]
$\Rightarrow B={{10}^{-7}}.\dfrac{10}{0.1}={{10}^{-5}}Wb{{m}^{-2}}$

Now that we have found the magnetic field around the electron, we can proceed with the procedure of finding a force around a moving charged particle in a magnetic field. The force on a charged particle is given by:
$F=qvB$
According to question q is the charge on electron, v is the velocity with which electron is moving and B we have calculated which came out to be ${{10}^{-5}}Wb{{m}^{-2}}$. Putting all the required values in the formula,
$F=(1.6\times {{10}^{-19}})({{10}^{6}})({{10}^{-5}})$
\[\therefore F=8\times {{10}^{-18}}\,N\]

Hence, the correct option is C.

Note: The thing to note here is that there cannot be a force on a stationary particle and without a magnetic field. The moving current produces a magnetic field due to which a force is generated on the moving particle. If there is no magnetic field then there will be no force on the particle.