Question

The current in the winding on a toroid is $2$A. there are $400$ turns and the mean circumferential length is $40cm$ with the aid of a search coil and charge measuring instrument, the magnetic field is found to be $1.0T$, the magnetization is

• A. $397$
• B. $200$
• C. $7.94\times {10}^5$
• D. $5\times {10}^5$

Answer 1

The correct option is A $397$

$Given:$ $I=2A$

$\text{No. of turns}=400turns$

$\text{circumferential length}=400cm$

$Solution:$ $n_{As,n}=\frac{400}{2\pi R}=\frac{400}{40\times {10}^{-2}}=1000$

$\mu =ni=1000\times 2=2000$

$\Rightarrow {\mu }_0{\mu }_r=\frac{1}{2000}=5\times {10}^{-4}$

$\Rightarrow {\mu }_r=\frac{5\times {10}^{-4}}{{\mu }_0}=\frac{5\times {10}^{-4}}{4\pi \times {10}^{-7}}=398$

$\Rightarrow \chi ={\mu }_r-1=397$

$So,the$ $correct$ $option:A$