A) 3 B
B) 5 B
C) 4 B
D) 2 B
Correct Answer: C
Solution :
Magnetic field at the centre of a current carrying loop is given by \[B=\frac{{{\mu }_{0}}\,ni}{2r}\] Here: \[n=\] no. of turns in loop \[i=\] current, \[{{r}_{1}}=\] radius of loop, \[{{r}_{1}}=r\] For \[n=1\] turn \[B=\frac{{{\mu }_{0}}i}{2{{r}_{1}}}\] ... (1) When \[n=2\] turns and radius \[{{r}_{2}}=\frac{r}{2},\,\,{{i}_{2}}=i\] or \[{{B}_{2}}=\frac{2{{\mu }_{0}}i\times 2}{2r}\] ?. (2) Now, from eqn. (1) and (2) \[\frac{{{B}_{2}}}{B}=4\] Hence, \[{{B}_{2}}=4B\]You need to login to perform this action.
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