Question: Q. 7. (i) How is a toroid different from a solenoid ?

(ii) Use Ampere’s circuital law to obtain the magnetic field inside a toroid.

(iii) Show that in an ideal toroid, the magnetic field (a) inside the toroid and (b) outside the toroid at any point in the open space is zero.

U] [O.D. Comptt. I, II, III 2014]

Show Answer

Solution:

Ans. (i) A toroid can be viewed as a solenoid which has been bent into a circular shape to close on itself. 1 (ii)

For the magnetic field at a point $\mathrm{S}$ inside a toroid we have

$$ \begin{aligned} B(2 \pi r) & =\mu_{0} N I \ B & =\mu_{0} \frac{N I}{2 \pi r}=\mu_{0} n I \end{aligned} $$

( $n=$ no. of turns per unit length of solenoid) $1 / 2$

$$ n=\frac{N}{2 \pi r} $$

(iii) For the loop 1, Ampere’s circuital law gives $B_{1} .2 \pi r_{1}=\mu_{0}(0)$, i.e. $B_{1}=0$

Thus the magnetic field, in the open space inside the toroid is zero

Also at point $Q$, we have $B_{3}\left(2 \pi r_{3}\right)=\mu_{0}\left(I_{\text {enclosed }}\right)$ But from the section cut, we see that the current coming out of the plane of the paper, is cancelled exactly by the current going into it.

Hence,

$$ I_{\text {enclosed }}=0 $$

$B_{3}=0$

$1 / 2$

[CBSE Marking Scheme 2014]

Answering Tips

  • Practice the derivation of Ampere’s Circuital Law to obtain the magnetic field inside a toroid.


विषयसूची

जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक