A) A quarter of its first value
B) Unaltered
C) Four times of its first value
D) A half of its first value
Correct Answer: C
Solution :
Magnetic field at the centre of current carrying coil is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi Ni}{r}\] Þ \[B\propto \frac{N}{r}\] Þ \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{N}_{1}}}{{{N}_{ & 2}}}\times \frac{{{r}_{2}}}{{{r}_{1}}}\]. The following figure shows that single turn coil changes to double turn coil. N1 = 1 N2 = 2 r1 = r r2 = r / 2 B1 = B B2 =? Þ \[\frac{B}{{{B}_{2}}}=\frac{1}{2}\times \frac{r/2}{r}=\frac{1}{4}\] Þ \[{{B}_{2}}=4B\] Short trick: For such type of problems remember \[{{B}_{2}}={{n}^{2}}{{B}_{1}}\]You need to login to perform this action.
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