question_answer

A length L of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is                       [NCERT 1980; AIIMS 1980; MP PMT 1995, 99]

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 \[{{B}_{2}}={{n}^{2}}{{B}_{1}}\]