A solenoid is 1 m long and 4 cm in diameter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$. Total number of turns $N = 5 	imes 1000 = 5000$. C

Vedclass · Accepted Answer

$4.396 	imes 10^{-2} \ T$

Vedclass · Answer

(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$. Total number of turns $N = 5 	imes 1000 = 5000$. Current $I = 7 \ A$. The magnetic field at the centre of a finite solenoid is given by $B = \frac{\mu_0 N I}{2\ell} (\cos 	heta_1 + \cos 	heta_2)$. Since the solenoid is long compared to its radius,we can approximate the field using the formula for an ideal solenoid $B = \mu_0 n I$,where $n = N/\ell$. $B = \frac{\mu_0 N I}{\ell} = \frac{4\pi 	imes 10^{-7} 	imes 5000 	imes 7}{1}$. $B = 4\pi 	imes 10^{-7} 	imes 35000 = 14\pi 	imes 10^{-3} \ T$. $B \approx 14 	imes 3.14159 	imes 10^{-3} \ T = 43.98 	imes 10^{-3} \ T = 4.398 	imes 10^{-2} \ T$. Rounding to the given options,the correct value is $4.396 	imes 10^{-2} \ T$.

$A$ solenoid is $1 \ m$ long and $4 \ cm$ in diameter. It has five layers of windings of $1000$ turns each and carries a current of $7 \ A$. The magnetic field at the centre of the solenoid is

Similar Questions

$A$ toroid core has an inner radius of $0.24 \ m$ and an outer radius of $0.26 \ m$. $A$ current of $10 \ A$ flows through the wire having $2500$ turns around it. Find the magnetic field inside the core of the toroid.

$A$ current $i$ ampere flows along the inner conductor of a coaxial cable and returns along the outer conductor of the cable. The magnetic induction at any point outside the conductor at a distance $r$ metre from the axis is

$A$ 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 ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

The region inside a current-carrying toroid is filled with a material (Niobium) having magnetic susceptibility $\chi = 2.6 \times 10^{-5}$. The percentage increase in the magnetic field in the presence of Niobium over that without it is:

$A$ long solenoid has $200$ turns per cm and carries a current of $2.5 \, A$. The magnetic field at its centre is (given $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module