Search any question & find its solution
Question:
Answered & Verified by Expert
The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$. The percentage increase in the value of magnetic field inside the toroid will be
Options:
Solution:
2076 Upvotes
Verified Answer
The correct answer is:
$2 \%$
Given,
Susceptibility of material, $\chi_m=2 \times 10^{-2}$
Using $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}=1+0.02=1.02$
$\mathrm{B}_{\text {final }}=\mu_{\mathrm{r}} \mathrm{B}_0$ (here, $\mathrm{B}_0=$ initial magnetic field)
$\%$ increase in magnetic field
$=\frac{\mathrm{B}_{\text {final }}-\mathrm{B}_0}{\mathrm{~B}_0} \times 100=\frac{\mu_{\mathrm{r}} \mathrm{B}_0-\mathrm{B}_0 \times 100}{\mathrm{~B}_0}$
$=\frac{(\chi+1)-1 \times 100}{1}=0.02 \times 100=2 \%$
Susceptibility of material, $\chi_m=2 \times 10^{-2}$
Using $\mu_{\mathrm{r}}=1+\chi_{\mathrm{m}}=1+0.02=1.02$
$\mathrm{B}_{\text {final }}=\mu_{\mathrm{r}} \mathrm{B}_0$ (here, $\mathrm{B}_0=$ initial magnetic field)
$\%$ increase in magnetic field
$=\frac{\mathrm{B}_{\text {final }}-\mathrm{B}_0}{\mathrm{~B}_0} \times 100=\frac{\mu_{\mathrm{r}} \mathrm{B}_0-\mathrm{B}_0 \times 100}{\mathrm{~B}_0}$
$=\frac{(\chi+1)-1 \times 100}{1}=0.02 \times 100=2 \%$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.