| In the figure shown a conducting wire PQ of length \[l\,=\,1\,m\], is moved in a uniform magnetic field \[B\,=4T\] with constant velocity \[v\,=\,2\,m/s\]towards right. |
| Given: \[R=2\Omega \], \[C=1\,\]\[F\] and \[L=4H\] |
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| Currents through resistance, capacitor and inductor at any time t are\[{{l}_{1}}\],\[{{l}_{2}}\] and \[{{l}_{3}}\]respectively, Current through wire PQ is l. Find the force required to move the wire with the given constant velocity of\[2m/s\] at \[t=2s\]: |
A) 8 N
B) 16 N
C) 24 N
D) 32 N
Correct Answer: D
Solution :
| \[{{F}_{applied}}(rightwards)={{F}_{m}}(leftwards)\] |
| \[=llB=\left( {{l}_{1}}+{{l}_{2}}+{{l}_{3}} \right)lB\] |
| \[=llB=\left( {{l}_{1}}+{{l}_{2}}+{{l}_{3}} \right)lB\] |
| \[{{l}_{3}}\] we have already calculated at 2 which is 4A. |
| \[{{l}_{2}}=\frac{dq}{dt}=\frac{d}{dt}(C{{V}_{C}})=\frac{d}{dt}(8)=0\] |
| And \[{{l}_{1}}=\frac{{{V}_{R}}}{R}=\frac{8}{2}=4A\] |
| \[{{F}_{applied}}=(4+4)(1)=32N\] |
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