A copper wire having a resistance of $$0.01$$ ohm in each metre is used to wind a $$400$$ turn solenoid of radius $$1.0\ cm$$ and length $$20\ cm$$. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of $$1.0 \times 10^{-2}$$ T near the centre of the solenoid.
As shown in the figure,radius of each coil=$$1cm$$
Number of turn=$$400$$
Total length of winding =N(perimeter of one coil)
$$=400(\pi(1\times 10^{-2})^{2})$$
$$=400\pi \times 10^{-4}=4\pi \times 10^{-2} m$$
Resistance of winding =$$1 \times $$length of winding coils
=$$1 \times (4\pi \times 10^{-2}) ohm$$
External battery cause a magnetic field .
Let us consider,current flowing in coil is $$i$$.
Magnetic field due to current at center O.
$$B=N\dfrac{\mu_0 l}{2r}$$
$$1.0\times 10^{-}2=N\dfrac{\mu_0 l}{2r}$$
$$ I=\dfrac { 2\times 1\times { 10 }^{ -2 }\times { 10 }^{ -2 } }{ 4\pi\times { 10 }^{ -7 }\times N } =\dfrac { 2\times { 10 }^{ 3 } }{ 4\pi N } $$
$$ I=\dfrac { 159.23 }{ N } A$$
$$ I=0.4A$$
Now circuit is shown in fig.3,
$$V=iR$$
$$=0.4\times4\pi \times 10^{-2}$$
$$=5\times 10^{-2} V$$
$$V=50mV$$
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