A solenoid is 1.0 m long and it has 4250 tur | vedclass

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

Vedclass · Accepted Answer

$2.7 	imes 10^{-2} \ Wb/m^2$

Vedclass · Answer

(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula: $B = \mu_0 n I$. Here, $n$ is the number of turns per unit length, which is calculated as $n = N/L$. Given: $N = 4250$, $L = 1.0 \ m$, $I = 5.0 \ A$, and $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$. First, calculate $n$: $n = 4250 / 1.0 = 4250 \ turns/m$. Now, substitute the values into the formula: $B = (4\pi 	imes 10^{-7}) 	imes 4250 	imes 5.0$. $B = (4 	imes 3.14159 	imes 10^{-7}) 	imes 21250$. $B \approx 12.566 	imes 10^{-7} 	imes 21250 \approx 0.0267 \ T$. Thus, $B \approx 2.7 	imes 10^{-2} \ Wb/m^2$.

$A$ solenoid is $1.0 \ m$ long and it has $4250$ turns. If a current of $5.0 \ A$ is flowing through it, what is the magnetic field at its centre? $[\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A]$

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