A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0.1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct valu

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Q: 72 (NDA-II/2018)
A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0.1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct value for the magnetic induction at the centre of the coil?

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,17,29,13,13,17,3

keywords: 

{'circular coil': [0, 0, 0, 2], 'magnetic induction': [0, 0, 0, 1], 'coil': [0, 0, 1, 1], 'radius': [0, 0, 2, 2], 'correct value': [0, 0, 0, 3], 'turns': [0, 0, 0, 2], 'centre': [10, 3, 7, 19], 'tesla': [0, 0, 1, 2]}

When the number of turns in the circular coil is doubled, the strength of the magnetic field at the center of the coil will also double. This is because the magnetic field produced by each turn of the coil adds up to the total magnetic field at the center. So, increasing the number of turns adds more magnetic field contributions, resulting in a higher magnetic field strength.

On the other hand, when the radius of the circular coil is halved, the strength of the magnetic field at the center will decrease by a factor of four. This is because the magnetic field is inversely proportional to the distance from the source (or the radius of the coil in this case). So, reducing the radius by half will result in four times less magnetic field strength.

Combining these effects, doubling the number of turns while halving the radius will result in a net increase in the magnetic field strength. Taking into account the given value of 0.1 tesla for the original coil, the correct value for the magnetic induction at the center of the coil after doubling the turns and halving the radius will be 0.4 tesla, as stated in option 3.