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

Q: 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

c (c) Magnetic field at the centre of current carrying coil is given by $B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2\pi Ni}}{r}$ $==>$ $B \propto \frac{N}{r}$ $==>$ $\frac{{{B_1}}}{{{B_2}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{r_2}}}{{{r_1}}}$. The following figure shows that single turn coil changes to double turn coil. $N_1 = 1 \,\,\,N_2 = 2$ $r_1 = r \,\,\,r_2 = r / 2$ $B_1 = B \,\,\,B_2 = ?$ $==>$ $\frac{B}{{{B_2}}} = \frac{1}{2} \times \frac{{r/2}}{r} = \frac{1}{4}$ $==>$ ${B_2} = 4B$ Short trick : For such type of problems remember ${B_2} = {n^2}{B_1}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A
A quarter of its first value
B
Unaltered
C
Four times of its first value
D
A half of its first value

AIIMS 1980, Medium

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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