Two very long,straight,and parallel wires car | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the two wires be placed in the $xy$-plane,parallel to the $y$-axis,at $x = -d/2$ and $x = d/2$. The currents are in opposite directions (e.g.,

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let the two wires be placed in the $xy$-plane,parallel to the $y$-axis,at $x = -d/2$ and $x = d/2$. The currents are in opposite directions (e.g.,$+I$ in the $y$-direction and $-I$ in the $y$-direction). At the midpoint (origin),the magnetic field $B_1$ due to the first wire is directed into the plane ($-z$ direction),and the magnetic field $B_2$ due to the second wire is also directed into the plane ($-z$ direction). The resultant magnetic field $B = B_1 + B_2$ is perpendicular to the plane of the wires (along the $-z$ direction). The velocity $v$ of the charge $q$ is given as perpendicular to the plane of the wires (along the $z$ direction). The magnetic force on a moving charge is given by $F = q(v 	imes B)$. Since the velocity vector $v$ and the magnetic field vector $B$ are collinear (both along the $z$-axis),the cross product $v 	imes B = 0$. Therefore,the magnitude of the magnetic force acting on the charge is $0$.

Similar Questions

An electron is projected along the axis of a circular conductor carrying some current. The electron will experience a force:

$A$ proton and an alpha particle moving with equal speeds enter normally into a uniform magnetic field. The ratio of times taken by the proton and the alpha particle to make one complete revolution in the magnetic field is

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module