A long copper wire is wound in the form of a coil of radius r. A current of 2 A is passed through this coil and the magnetic induction at the centre of this coil is noted. The same wire is now folded end to end and coil of the same radius r is prepared and the same current is passed through it. The magnetic induction at the centre

Will be doubled

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Will be halved

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Will remain same

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Will drop to zero

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Solution

The correct option is B Will be halved
Case- $I$
magnetic field ddue to circular coil at $0$ is $B$ .
$B_{1} = \frac{μ_{0} I}{2 r}$
Case- $I I$
Magnetic Induction at $0$ is
$B_{2} = \frac{1}{2} \times \frac{μ_{0} I}{2 r}$
Hence, clearly $B_{2} = \frac{1}{2} \times B_{1}$

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The correct option is B Will be halvedCase-Imagnetic field ddue to circular coil at 0 is B.B1=μ0I2rCase-IIMagnetic Induction at 0 is B2=12×μ0I2rHence, clearly B2=12×B1

The correct option is B Will be halved
Case- $I$
magnetic field ddue to circular coil at $0$ is $B$ .
$B_{1} = \frac{μ_{0} I}{2 r}$
Case- $I I$
Magnetic Induction at $0$ is
$B_{2} = \frac{1}{2} \times \frac{μ_{0} I}{2 r}$
Hence, clearly $B_{2} = \frac{1}{2} \times B_{1}$