A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be:

2 : 1

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1 : 4

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4 : 1

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1 : 2

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Solution

The correct option is B $2 : 1$
Let the radii be $r_{1}$ and $r_{2}$ respectively.
Since there are two turns of radius $r_{2}$ , $r_{1} = 2 r_{2}$
Magnetic field B at the center of the coil of radius $r_{1}$ $B_{1} = \frac{μ_{o} i}{2 r_{1}} = \frac{μ_{o} i}{4 r_{2}}$
Magnetic field B at the center of the coil of radius $r_{2}$ $B_{2} = 2 \times \frac{μ_{o} i}{2 r_{2}}$
$∴ \frac{B_{2}}{B_{1}} = \frac{2 \times \frac{μ_{o} i}{2 r_{2}}}{\frac{μ_{o} i}{4 r_{2}}} = 4$
$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{1}{4}$

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A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be:

The correct option is B 2:1Let the radii be r1 and r2 respectively.Since there are two turns of radius r2, r1=2r2Magnetic field B at the center of the coil of radius r1 B1=μoi2r1=μoi4r2Magnetic field B at the center of the coil of radius r2 B2=2×μoi2r2∴B2B1=2×μoi2r2μoi4r2=4⇒B1B2=14