Question

FIGURE shows a U-shaped conducting rail that is oriented vertically in a horizontal magnetic field. The rail has no electric resistance and does not move. A slide wire with mass $m$and resistance $R$can slide up and down without friction while maintaining electrical contact with the rail. The slide wire is released from rest.

a. Show that the slide wire reaches a terminal speed $v_{\text{term}}$, and find an expression for$v_{\text{term}}$ .

b. Determine the value of $v_{\text{term}}$if $l=20\mathrm{cm},m=10\mathrm{g},R=$$0.10\Omega$and $B=0.50\mathrm{T}$

Short answer

(a) Expression of Terminal velocity, $v_{\text{term}}=\frac{mgR}{B^2{l}^2}$

(b) Value of Terminal velocity,$v_{\text{term}}=0.98\mathrm{m}/\mathrm{s}$

Step-by-step solution

Find expression for Vterm  (part a)

The rod is determined by two forces: an upward magnetic force and a downward weight force. When the two forces are balanced, the velocity becomes constant.

$F_B=mg$

When a current-carrying wire moves through a magnetism, the field imposes a magnetic pull on the wire. The magnetic force exerted by a wire of length is described by an equation.

$F_B=IlB$

Using Faraday's law

Using Faraday's law to get the magnitude of emf in terms of moving antenna using equation,

$\epsilon =Blv$

Using Ohm's law to get the Induced current

$I=\frac{\epsilon }{R}=\frac{Blv}{R}$

Use this expression to get the magnetic force,

$F_B=IlB=\left(\frac{Blv}{R}\right)lB=\frac{B^2{l}^{2}v}{R}$

As $F_B=mg$, so we use equation (4) to get the terminal velocity by$mg=\frac{B^2{l}^2{v}_{\text{term}}}{R}$

$v_{\text{term}}=\frac{mgR}{B^2{l}^2}$

Find Terminal velocity (part b)

$v_{\text{term}}=\frac{mgR}{B^2{l}^2}$

$=\frac{\left(10\times {10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(0.10\Omega )}{(0.50\mathrm{T}{)}^2(0.20\mathrm{m}{)}^2}$

$=0.98\mathrm{m}/\mathrm{s}$