Consider a tightly wound 100 turn coil of rad | vedclass

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

Question

(C) Given:Number of turns,$n = 100$Radius,$r = 10 \,cm = 0.1 \,m$Current,$I = 2 \,A$Permeability of free space,$\mu_0 = 4\pi 	imes 10^{-7} \,T \cdot

Vedclass · Accepted Answer

$12.56 	imes 10^{-4} \,T$

Vedclass · Answer

(C) Given:Number of turns,$n = 100$Radius,$r = 10 \,cm = 0.1 \,m$Current,$I = 2 \,A$Permeability of free space,$\mu_0 = 4\pi 	imes 10^{-7} \,T \cdot m/A$The magnetic field $B$ at the centre of a circular coil is given by the formula:$B = \frac{\mu_0 n I}{2r}$Substituting the values:$B = \frac{(4\pi 	imes 10^{-7}) 	imes 100 	imes 2}{2 	imes 0.1}$$B = \frac{4\pi 	imes 10^{-5} 	imes 2}{0.2}$$B = \frac{8\pi 	imes 10^{-5}}{0.2} = 40\pi 	imes 10^{-5} = 4\pi 	imes 10^{-4} \,T$Using $\pi \approx 3.14$:$B = 4 	imes 3.14 	imes 10^{-4} \,T = 12.56 	imes 10^{-4} \,T$

Consider a tightly wound $100$ turn coil of radius $10 \,cm$ carrying a current of $2 \,A$. The magnitude of the magnetic field at the centre of the coil is

Similar Questions

$A$ straight wire carrying a current of $12 \,A$ is bent into a semi-circular arc of radius $2 \,cm$ as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is

Among the following,the unit of permeability is $NOT$ represented by

$20 \, A$ current is flowing in a long straight wire. The intensity of the magnetic field at a distance of $20 \, cm$ from the wire will be:

$A$ conducting wire carrying current $I$ is arranged as shown in the figure. Find the magnetic field at point $O$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module