A 50 cm long solenoid has a winding of 400 t | vedclass

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(B) The magnetic field at the center of a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length $(n = N/L)$.Give

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$12.5$

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(B) The magnetic field at the center of a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length $(n = N/L)$.Given:Length $L = 50 \ cm = 0.5 \ m$Number of turns $N = 400$Magnetic field $B = 4 \pi 	imes 10^{-3} \ T$Permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$First, calculate the number of turns per unit length:$n = \frac{N}{L} = \frac{400}{0.5} = 800 \ turns/m$Now, substitute the values into the formula $B = \mu_0 n I$:$4 \pi 	imes 10^{-3} = (4 \pi 	imes 10^{-7}) 	imes 800 	imes I$$I = \frac{4 \pi 	imes 10^{-3}}{4 \pi 	imes 10^{-7} 	imes 800}$$I = \frac{10^{-3}}{10^{-7} 	imes 800} = \frac{10^4}{800} = \frac{100}{8} = 12.5 \ A$.

$A$ $50 \ cm$ long solenoid has a winding of $400$ turns. What current must pass through it to produce a magnetic field of induction $4 \pi \times 10^{-3} \ T$ at the centre (in $A$)?

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