A long solenoid is formed by winding 20 turns | vedclass

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(A) The magnetic field $B$ inside 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

Vedclass · Accepted Answer

$8$

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(A) The magnetic field $B$ inside 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.Given:$B = 20 	ext{ mT} = 20 	imes 10^{-3} 	ext{ T}$$n = 20 	ext{ turns/cm} = 20 	imes 100 	ext{ turns/m} = 2000 	ext{ turns/m}$$\frac{\mu_0}{4\pi} = 10^{-7} 	ext{ T m/A} \Rightarrow \mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$Rearranging the formula for current $i$:$i = \frac{B}{\mu_0 n}$Substituting the values:$i = \frac{20 	imes 10^{-3}}{4\pi 	imes 10^{-7} 	imes 2000}$$i = \frac{20 	imes 10^{-3}}{8\pi 	imes 10^{-4}}$$i = \frac{200}{8\pi} = \frac{25}{\pi} \approx \frac{25}{3.14} \approx 7.96 	ext{ A}$Rounding to the nearest integer,we get $i \approx 8 	ext{ A}$.

$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

Similar Questions

$A$ closely wound solenoid of length $1 \,m$ has $5$ layers of $500$ turns each. If the magnitude of the magnetic field inside the solenoid near its centre is $4.4 \,mT$, the current carried is: (in $\,A$)

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The magnetic field inside a long solenoid is:

Inductance per unit length near the middle of a long solenoid is ($\mu_0=$ permeability of free space,$n=$ number of turns per unit length,$d=$ the diameter of the solenoid).

An infinitely long cylindrical wire of radius $a$ carries a steady current $I$ uniformly distributed across its cross-section. Determine the magnetic field $B$ at a distance $r$ from the axis for: $(a) r > a$,$(b) r = a$,$(c) r < a$,and $(d)$ at the axis $(r = 0)$.

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