A) \[x/8\]
B) \[x/4\]
C) \[x/2\]
D) \[2x\]
Correct Answer: A
Solution :
The magnetic field at the centre of a current carrying loop is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{2\pi i}{a} \right)=\frac{{{\mu }_{0}}i}{2a}\] The magnetic moment at the centre of current carrying loop is given by \[M=i(\pi {{a}^{2}})\] Thus, \[\frac{B}{M}=\frac{{{\mu }_{0}}i}{2a}\times \frac{1}{i\pi {{a}^{2}}}=\frac{{{\mu }_{0}}}{2\pi {{a}^{3}}}=x\,(given)\] When both the current and the radius are doubled, the ratio becomes \[\frac{{{\mu }_{0}}}{2\pi {{(2a)}^{3}}}=\frac{{{\mu }_{0}}}{8(2\pi {{a}^{3}})}=\frac{x}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
