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