A) \[1.256 \times 1{{0}^{-}}^{3}\,T\]
B) \[1.679 \times 1{{0}^{-}}^{5}\,T\]
C) \[1.512 \times 1{{0}^{-}}^{5}T\]
D) \[2.28 \times 1{{0}^{-}}^{4}\,T\]
Correct Answer: A
Solution :
The magnetic field at the centre of the coil is \[B=\frac{{{\mu }_{0}}}{4\pi }\,\frac{2\pi Ni}{a}\] where N is number of turns, i is current and a is radius of coil. Given, \[\operatorname{N}= 25,\,\,D =10 cm= 10 \times 1{{0}^{-}}^{2}m\] \[a=\frac{D}{2}\,\,=\,\,\frac{10\times {{10}^{-2}}}{2}\,m\,\,=\,\,0.05\,m\] \[i\,\,=\,\,4\,A\] \[\therefore \,\,\,\,\,\,B={{10}^{-7}}\,\times \frac{2\times 3.14\times 25\times 4}{0.05}\] \[B=1.256\,\,\,\times \,\,{{10}^{\text{-3}}}T\]
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