A) \[250\mu T\]
B) \[150\mu T\]
C) \[125\mu T\]
D) \[75\mu T\]
Correct Answer: A
Solution :
The magnetic field at a point on the axis of a circular loop at a distance\[x\]from the centre is \[B=\frac{{{\mu }_{0}}i{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\] ?(i) Given: \[B=54\text{ }\mu T,\text{ }x=4\text{ }cm,\text{ }R=3\text{ }cm\] Putting the given values in Eq. (i), we get \[\therefore \] \[54=\frac{{{\mu }_{0}}i\times {{(3)}^{2}}}{2{{({{3}^{2}}+{{4}^{2}})}^{3/2}}}\] \[\Rightarrow \] \[54=\frac{9{{\mu }_{0}}i}{2{{(25)}^{3/2}}}=\frac{9{{\mu }_{0}}i}{2\times {{(5)}^{3}}}\] \[\therefore \] \[{{\mu }_{0}}i=\frac{54\times 2\times 125}{9}\] \[{{\mu }_{0}}i=1500\,\mu T-cm\] ?(ii) Now, putting\[x=0\]in Eq. (i), magnetic field at the centre of loop is \[B=\frac{{{\mu }_{0}}i{{R}^{2}}}{2{{R}^{3}}}=\frac{{{\mu }_{0}}i}{2R}=\frac{1500}{2\times 3}\] \[=250\text{ }\mu T\] [from Eq. (ii)]You need to login to perform this action.
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