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 inductions at their centres will be

Question

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 inductions at their centres will be (A) 4: 1 (B) 1: 4 (C) 2: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

Magnetic field at the centre of the coil,

B = \frac{μ _{0}}{2 π} \frac{N I}{a}

Let

l

be the length of the wire, then

B_{1} = \frac{μ _{0}}{2 π} \cdot \frac{1 \times I}{l /2 π} = \frac{μ _{0} I}{l}

and

B_{2} = \frac{μ _{0}}{2 π} \cdot \frac{2 \times I}{l /4 π} = \frac{4 μ _{0} I}{l}

Therefore,

\frac{B _{1}}{B _{2}} = \frac{1}{4}

or,

B_{1} : B_{2} = 1 : 4

.

Q. 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 inductions at their centres will be

4 : 1

1 : 4

2 : 1

1 : 2

Q. 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 inductions at their centres will be

4 : 1

1 : 4

2 : 1

1 : 2

Magnetic field at the centre of the coil, B=2πμ0​​aNI​ Let l be the length of the wire, then B1​=2πμ0​​⋅l/2π1×I​=lμ0​I​ and B2​=2πμ0​​⋅l/4π2×I​=l4μ0​I​ Therefore, B2​B1​​=41​ or, B1​:B2​=1:4.