Question

A rectangular frame of wire abcd has dimensions $32cm\times 8.0$ cm and a total resistance of $2.0\Omega .$ It is pulled out of a magnetic field B = 0.020 T by applying a force of $3.2\times {10}^{-5}$ N figure . It is found that the frame moves with constant speed. Find

(a) This constant speed

(b) The emf induced in the loop

(c) The potential difference between the points a and b and

(d) The potential difference between the points c and d

Answer 1

R = $2.0\Omega$,

B = 0.020 T,

l = 32 cm = 0.32 m,

b = 8 cm = 0.08 m

(a) F = ilB = $3.2\times {10}^{-5}N$

$\Rightarrow \frac{B^2{l}^{2}v}{R}=3.2\times {10}^{-5}$

$\Rightarrow \frac{(0.020{)}^2\times (0.08{)}^2\times v}{2}=3.2\times {10}^{-5}$

$\Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}}$

(b) emf = vBl = $2.5\times 0.02\times 0.08$

= $4\times {10}^{-2}V$

(c) Resistance per unit length

= $\frac{2}{0.8}$

Resistance of part $\frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\Omega$

$v_{ab}=iR=\frac{Blv}{2}\times 1.8$

=$\frac{0.2\times 0.08\times 25\times 1.8}{2}$

= $0.036V=3.6\times {10}^{-2}V$

(d) Resistance of cd

= $\frac{2\times 0.8}{0.8}0.2\Omega$

V = iR = $\frac{2\times 0.08\times 25\times 0.2}{2}$

= $4\times {10}^{-3}V.$