Question

A rectangular frame of wire abcd has dimensions $32cm\times 8.0cm$ and a total resistance of 2.0$\Omega$. It is pulled out of a magnetic field B=0.02T 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

Hint: For part a: Use the formula for magnetic force. Substitute current in formula for magnetic force in terms of velocity. Then substitute the values and calculate velocity. For part b: Use the formula for induced EMF substitute the values and get emf induced in the loop. For part c and d: Find the resistance per unit length. And then calculate resistance from a to b and c to d. Then calculate potential differences across them.

Formula used:
$F=ilB $
$\epsilon=vBl$
$i= \dfrac {Blv}{R}$

Complete answer:
Given: R=$2\Omega$
            B= 0.02T
            l= 32 cm=0.32m
            b= 8cm=0.08m

a.)Let the speed of the frame be v,
Magnetic force is given by,
F=ilB = $3.2\times { 10 }^{ -5 }N$ ..(1)
But, current in the coil I is given by,
$i= \dfrac {Blv}{R}$ …(2)
Substituting, equation, (2) in equation. (1) we get,
${ B }^{ 2 }{ l }^{ 2 }\dfrac { v }{ R }=3.2\times { 10 }^{ -5 }N$
Now, substituting values in above equation we get,
${ 0.02 }^{ 2 }\times{ 0.32 }^{ 2 }\times \dfrac { v }{ 2 }=3.2\times { 10 }^{ -5 }N$
$0.0004\times 0.1024 \times \dfrac { v }{ 2 }=3.2\times { 10 }^{ -5 }N$
$\Rightarrow v=\cfrac { 2\times 3.2\times { 10 }^{ -5 } }{ 0.0004\times 0.1024 }$
$\Rightarrow v=25 {m}/{s}$
Hence, the speed is $25 {m}/{s}$.

b.)EMF induced is given by,
$\epsilon=vBl$
Substituting values in above equation we get,
$\epsilon= 25 \times 0.02 \times 0.08$
$\Rightarrow \epsilon= 0.04 V$
Thus, the EMF induced in the loop is 0.04V.

c.)Resistance per unit length is given by,
$r= \dfrac {0.2}{0.8}$ …(Since,0.32+0.32+0.08+0.08=0.8)
Now, ratio of resistance of part,
$\dfrac {ab}{cd}= \dfrac {2 \times 0.72}{0.8}$
$\dfrac {ab}{cd}= 1.8\Omega$

Potential difference across points a and b is given by,
${V}_{ab}= iR$
But, $i= \dfrac {Blv}{R}$
$\therefore{V}_{ab}= \dfrac {Blv}{2} \times1.8$
Substituting the values in above equation we get,
${V}_{ab}= \dfrac {0.02 \times 0.08 \times 25 \times 1.8}{2}$
${V}_{ab}=0.036 V$
Thus, the potential difference across points a and b is 0.036V.

d.)Resistance in cd= $\dfrac {2 \times 0.08}{0.8}$
$\dfrac {ab}{cd}= 0.2\Omega$

Potential difference across points c and d is given by,
${V}_{cd}= iR$
But, $i= \dfrac {Blv}{R}$
$\therefore{V}_{ab}= \dfrac {Blv}{2} \times0.2$
Substituting the values in above equation we get,
${V}_{ab}= \dfrac {0.02 \times 0.08 \times 25 \times 0.2}{2}$
${V}_{ab}=0.004 V$
Thus, the potential difference across points a and b is 0.004V.

Note:
As the question looks lengthy so don't get panic. Solve the subparts one after the other. You need to memorize these formulas properly.
While calculating resistance per unit length, don’t just divide the total resistance by length of the frame but by addition of the lengths of all sides.