In Figure, a metal wire of mass m = 24.1 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 2.56 cm. The track lies in a vertical uniform magnetic field of magnitude 56.3 mT. At time t = 0, device Gis connected to the rails, producing a constant current i = 9.13 mA in the wire and rails (even as the wire moves). Att = 61.1 ms, (a) what is the wire’s speed? (b) What is the wire’s direction of motion (left or right)?

1. The mass of the metal wire, $m=24.1mgor24.1\times {10}^{-3}g$
2. The distance between two horizontal parallel rails, $d=2.56cmor2.56\times {10}^{-2}m$
3. The magnitude of the vertical magnetic field, $\left|B\right|=56.3mTor56.3\times {10}^{-3}T$
4. The constant current in the wire, localid="1662955985002">i=9.13mAor9.13×10-3A
5. The time of current flow,$t=61.1msor61.1\times {10}^{-3}s$

If a particle is moving with a uniform velocity within a uniform magnetic field, then it experiences a magnetic force due to its charge value that induces a current within the loop. The force acting on the particle is due to the current along the length of the conductive wire. The direction of this magnetic force is given by Fleming's right-hand rule as the force is perpendicular to both the speed and magnetic field acting on it.

The magnetic force along a loop wire is as follows:

$\stackrel{\rightharpoonup }{d{F}_B}=i\left(\stackrel{\rightharpoonup }{dL}\times \overrightarrow{B}\right)$ …… (i)

Here, i is the current in the loop, $\stackrel{\rightharpoonup }{B}$is the magnetic field vector that it experiences,$\stackrel{\rightharpoonup }{dL}$is the length vector of the conducting wire.

The acceleration of the charged particle in the conductor is as follows:

$a=\frac{v}{t}$ …… (ii)

Here, v is the velocity of the charged particle, t is the time taken for the motion.

The force according to Newton’s second law is as follows:

F = ma ….. (iii)

Here, m is the mass of the body, a is the acceleration of the body in motion.

Here, force is magnetic force; hence F = F_B

The speed of the wire is given using equations (ii) and (iii) as follows:

$v=\frac{F_B}{m}t$ (iv)

Here, the direction of magnetic field is perpendicular to the distance between two horizontal parallel rails, that is $\theta ={90}^o$. Thus, the force on the current carrying wire is given using equation (i) as follows:

$F_B=idB$

Here, d is the distance of the parallel long rails.

Now, the magnitude of the speed of the wire is given using the above value in equation (iv) as follows:

$\begin{array}{rcl}v& =& \frac{idB}{m}t\\ & =& \frac{\left(9.13\times {10}^{-3}A\right)\times \left(2.56\times {10}^{-2}m\right)\times \left(56.3\times {10}^{-3}T\right)\times \left(61.1\times {10}^{-3}s\right)}{24.1\times {10}^{-3}g}\\ & =& 3.34\times {10}^{-2}\frac{m}{s}\end{array}$

Hence, the value of the speed is $3.34\times {10}^{-2}\frac{m}{s}$.

According to the right hand rule, direction of motion of the wire is left. Hold the right hand in such way that the curled figures denote the direction of the rotation of the vectors and the upstretched thumb denote the direction of the vector product of two vectors.

Here, the length of the wire has the same direction as the current flowing through the wire. The force acts on the wire along the thumb towards left side.

Hence, the motion of the wire is towards the left side.