A) four times
B) same
C) half
D) double
Correct Answer: B
Solution :
When a circular coil of radius a, carrying current i, is kept in a magnetic field B, the magnetic field at the centre of the coil is \[B=\frac{{{\mu }_{0}}Ni}{2\,a}\] where N is number of turns in the coil Given, \[{{i}_{1}}=i,\,{{i}_{2}}=2i,\,{{N}_{1}}=N,\,{{N}_{2}}=\frac{N}{2}\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{N}_{1}}{{i}_{1}}}{{{N}_{2}}{{i}_{2}}}\] \[=\frac{N\times i}{(N/2)\times 2i}\] \[\Rightarrow \] \[{{B}_{1}}={{B}_{2}}\] Hence, magnetic field remains same.
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