In a toroid,the magnetic field on the axis is | vedclass

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

Question

(B) The magnetic field $B$ inside a toroid is given by the formula:$B = \mu_0 \mu_r n I$where $n = \frac{N}{2 \pi r}$ is the number of turns per unit

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The magnetic field $B$ inside a toroid is given by the formula:$B = \mu_0 \mu_r n I$where $n = \frac{N}{2 \pi r}$ is the number of turns per unit length.Substituting the values: $N = 250$,$r = 0.5 \, cm = 0.5 	imes 10^{-2} \, m$,$I = 1.5 \, A$,$\mu_r = 700$,and $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$.$B = (4 \pi 	imes 10^{-7}) 	imes 700 	imes \left[ \frac{250}{2 \pi 	imes 0.5 	imes 10^{-2}} ight] 	imes 1.5$$B = (2 	imes 10^{-7}) 	imes 700 	imes \left[ \frac{250}{0.5 	imes 10^{-2}} ight] 	imes 1.5$$B = (14 	imes 10^{-5}) 	imes (50000) 	imes 1.5$$B = 14 	imes 5 	imes 1.5 	imes 10^{-1} = 10.5 \, T$.

In a toroid,the magnetic field on the axis is to be calculated given the radius $r = 0.5 \, cm$,current $I = 1.5 \, A$,number of turns $N = 250$,and relative permeability $\mu_r = 700$. Find the magnetic field $B$ (in Tesla). (in $.5$)

Similar Questions

$A$ current of $1/4\pi \ A$ is flowing through a toroid. It has $1000$ turns per meter. The value of the magnetic field (in $Wb/m^2$) along its axis 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 $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$):

$A$ solenoid of length $50 \,cm$ having $100$ turns carries a current of $2.5 \,A$. The magnetic field at one end of the solenoid is

$A$ long straight wire of circular cross-section (radius $a$) is carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. The magnetic field is

State Ampere's circuital law and explain its mathematical expression.

Vedclass Test Series

Exam Paper Generator

Online Exam Module