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(B) The magnetic field at the centre of a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the cu

Vedclass · Accepted Answer

$1.05 	imes 10^{-2} \ Wb/m^2$

Vedclass · Answer

(B) The magnetic field at the centre of a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.For the first solenoid: $B_1 = \mu_0 n_1 i_1 = 6.28 	imes 10^{-2} \ Wb/m^2$,where $n_1 = 200 \ turns/cm$ and $i_1 = i$.For the second solenoid: $n_2 = 100 \ turns/cm$ and $i_2 = i/3$.Taking the ratio: $\frac{B_2}{B_1} = \frac{\mu_0 n_2 i_2}{\mu_0 n_1 i_1} = \frac{n_2 i_2}{n_1 i_1}$.Substituting the values: $\frac{B_2}{6.28 	imes 10^{-2}} = \frac{100 	imes (i/3)}{200 	imes i} = \frac{100}{200 	imes 3} = \frac{1}{6}$.Therefore,$B_2 = \frac{6.28 	imes 10^{-2}}{6} \approx 1.05 	imes 10^{-2} \ Wb/m^2$.

$A$ long solenoid has $200$ turns per $cm$ and carries a current $i$. The magnetic field at its centre is $6.28 \times 10^{-2} \ Wb/m^2$. Another long solenoid has $100$ turns per $cm$ and it carries a current $i/3$. The value of the magnetic field at its centre is:

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