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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.For the first solenoid: $n_1 = 200 ~turns/cm$,$i_1 = i$,and $B_1 = 6.28 	imes 10^{-2} ~Wb/m^2$.So,$6.28 	imes 10^{-2} = \mu_0 	imes 200 	imes i$  --- $(1)$For the second solenoid: $n_2 = 100 ~turns/cm$ and $i_2 = \frac{i}{3}$.The magnetic field $B_2$ is given by:$B_2 = \mu_0 	imes n_2 	imes i_2 = \mu_0 	imes 100 	imes \left(\frac{i}{3}ight)$Substituting the value of $i$ from equation $(1)$,where $i = \frac{6.28 	imes 10^{-2}}{\mu_0 	imes 200}$:$B_2 = \mu_0 	imes 100 	imes \left(\frac{1}{3} 	imes \frac{6.28 	imes 10^{-2}}{\mu_0 	imes 200}ight)$$B_2 = \frac{1}{3} 	imes \frac{100}{200} 	imes 6.28 	imes 10^{-2}$$B_2 = \frac{1}{3} 	imes \frac{1}{2} 	imes 6.28 	imes 10^{-2} = \frac{6.28 	imes 10^{-2}}{6} \approx 1.046 	imes 10^{-2} ~Wb/m^2$Rounding off,we get $B_2 \approx 1.05 	imes 10^{-2} ~Wb/m^2$.

$A$ 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 $\frac{i}{3}$. The value of the magnetic field at its centre is nearly:

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