A) independent of L
B) proportional to L2
C) inversely proportional to L
D) linearly proportional to L
Correct Answer: C
Solution :
Magnetic field at the centre due to either arm \[=\frac{2\sqrt{gh}}{g}=2\sqrt{\frac{h}{g}}\] \[=\frac{M{{R}^{2}}}{2}\] \[=M\left( \frac{{{L}^{2}}}{I2}+\frac{{{L}^{2}}}{4} \right)\] Field at centre due to the four arms of the square \[\frac{M{{L}^{2}}}{12}+\frac{M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}\] i.e., \[L=\sqrt{3}R\]
You need to login to perform this action.
You will be redirected in
3 sec
