A long straight wire AB carries a current of `4A`. A proton P travels at `4xx10^6ms^-1` parallel to the wire, `0*2m` from it and in a direction opposite to the current as shown in figure. Calculate the force which the magnetic field of current exerts on the proton. Also specify the direction of the force.

Options

A. `2.56xx10^(-18)N` towards left
B. `2.56xx10^(-18)N` towards right
C. `5.12xx10^(-18)N` towards left
D. `5.12xx10^(-8)N` towards right

Answer

B

Explanation

Magnetic field at P due to current `-` carrying wire AB is
`B=(mu_(0))/(4pi).(2I)/(a)=(10^(-7)xx2xx4)/(0.2)=4xx10^(-6)T`
By right `-` hand grip rule, the magnetic field at P is directed perpendicular to the plane of the paper and in inward direction.
Force on proton.
`F_(m)=qvB=(1.6xx10^(-19))xx(4xx10^(6))xx4xx10^(-6)=2.56xx10^(-18)N`
By right `-` hand rule for cross product, the force on the proton acts parallel to horizontal towards right.