A long solenoid of 50, cm length having 100 t | vedclass

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(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula: $B = \mu_{0} n I$,where $n = \frac{N}{\ell}$.Given:Length $\ell =

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$62.8$

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(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula: $B = \mu_{0} n I$,where $n = \frac{N}{\ell}$.Given:Length $\ell = 50\, cm = 0.5\, m$Number of turns $N = 100$Current $I = 2.5\, A$Permeability $\mu_{0} = 4\pi 	imes 10^{-7}\, T\, m\, A^{-1}$Substituting the values:$B = (4\pi 	imes 10^{-7}) 	imes \left(\frac{100}{0.5}ight) 	imes 2.5$$B = (4 	imes 3.14159 	imes 10^{-7}) 	imes 200 	imes 2.5$$B = 12.566 	imes 10^{-7} 	imes 500$$B = 6283.18 	imes 10^{-7} = 6.283 	imes 10^{-4}\, T$To express this in the form $...... 	imes 10^{-5}\, T$:$B = 62.83 	imes 10^{-5}\, T$Rounding to the nearest option,the value is $62.8$.

$A$ long solenoid of $50\, cm$ length having $100$ turns carries a current of $2.5\, A$. The magnetic field at the centre of the solenoid is $...... \times 10^{-5}\, T$. $(\mu_{0} = 4\pi \times 10^{-7}\, T\, m\, A^{-1})$

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