What is the net force on the square coil
Medium
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(a) Force on side $BC$ and $AD$ are equal but opposite so their net will be zero.
But ${F_{AB}} = {10^{ - 7}} \times \frac{{2 \times 2 \times 1}}{{2 \times {{10}^{ - 2}}}} \times 15 \times {10^{ - 2}} = 3 \times {10^{ - 6}}\,N$
and ${F_{CD}} = {10^{ - 7}} \times \frac{{2 \times 2 \times 1}}{{\left( {12 \times {{10}^{ - 2}}} \right)}} \times 15 \times {10^{ - 2}}$$ = 0.5 \times {10^{ - 6}}\,N$
$==>$ $\,{F_{net}} = {F_{AB}} - {F_{CD}}$ $ = 2.5 \times {10^{ - 6}}\,N$
$ = 25 \times {10^{ - 7}}\,N$, towards the wire.
But ${F_{AB}} = {10^{ - 7}} \times \frac{{2 \times 2 \times 1}}{{2 \times {{10}^{ - 2}}}} \times 15 \times {10^{ - 2}} = 3 \times {10^{ - 6}}\,N$
and ${F_{CD}} = {10^{ - 7}} \times \frac{{2 \times 2 \times 1}}{{\left( {12 \times {{10}^{ - 2}}} \right)}} \times 15 \times {10^{ - 2}}$$ = 0.5 \times {10^{ - 6}}\,N$
$==>$ $\,{F_{net}} = {F_{AB}} - {F_{CD}}$ $ = 2.5 \times {10^{ - 6}}\,N$
$ = 25 \times {10^{ - 7}}\,N$, towards the wire.
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