N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A enclos

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.

N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity $ω$ as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, $I_{A} - I_{B}$ , for the given Amperian loops is [2025]

1 $\frac{N^{2}}{2 π} q ω$

2 $\frac{2 π}{N} q ω$

3 $\frac{N}{2 π} q ω$

4 $\frac{N}{π} q ω$

Ans.
(3)

$I = \frac{N q}{\frac{2 π}{ω}} = \frac{N q ω}{2 π}$

Now for bigger Amperian loop (loop B)

$I_{B (enclosed)} = 0$

And for smaller Amperian loop (loop A)

$| I_{A (enclosed)} | = I = \frac{N q ω}{2 π}$

So, $| I_{A} - I_{B} | = \frac{N q ω}{2 π}$

Want Us to Email you About Special Offers & Updates?