A) 3B
B) 56
C) 4B
D) 2B
Correct Answer: C
Solution :
Magnetic field at the centre of a current carrying loop is given by \[B=\frac{{{\mu }_{0}}ni}{2{{r}_{1}}}\] 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 \[{{r}_{2}}=\frac{r}{2},\,\,{{i}_{2}}=i\] \[{{B}_{2}}=\frac{{{\mu }_{0}}\times 2\times i}{2\left( \frac{r}{2} \right)}\] 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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