The magnetic field at the centre of a long so | vedclass

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Question

$(A)$ 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

Vedclass · Accepted Answer

$1.56 	imes 10^{-2} \,T$

Vedclass · Answer

$(A)$ 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: $B_1 = \mu_0 n_1 i_1 = 6.24 	imes 10^{-2} \,T$,where $n_1 = 400$ and $i_1 = i$.So,$\mu_0 (400) i = 6.24 	imes 10^{-2} \,T$.For the second solenoid: $n_2 = 200$ and $i_2 = \frac{i}{2}$.The magnetic field $B_2 = \mu_0 n_2 i_2 = \mu_0 (200) \left( \frac{i}{2} ight) = \mu_0 (100) i$.Comparing $B_2$ with $B_1$: $B_2 = \frac{\mu_0 (100) i}{\mu_0 (400) i} 	imes B_1 = \frac{1}{4} 	imes 6.24 	imes 10^{-2} \,T$.$B_2 = 1.56 	imes 10^{-2} \,T$.

The magnetic field at the centre of a long solenoid having $400$ turns per unit length and carrying a current $i$ is $6.24 \times 10^{-2} \,T$. The magnetic field at the centre of another long solenoid having $200$ turns per unit length and carrying a current $\frac{i}{2}$ is

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