Magnetic Effects & Magnetism

JEE Main level, single correct, medium

58 of 276

Question 58 of 276

The electric current in a circular coil of 22 turns produces a magnetic induction B1{B}_{1} at its centre. The coil is unwound and is rewound into a circular coil of 55 turns and the same current produces a magnetic induction B2{B}_{2} at its centre.
The ratio of B2B1\dfrac{{B}_{2}}{{B}_{1}} is:

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Answer: (B) 254\dfrac{25}{4}

Magnetic field on the centre of a circular current coil is given by B=Nμ0i2RB=\dfrac{N{\mu }_{0}i}{2R}, here, NN is number of turns, ii is current and RR is radius of loop.
Now, for, N=2N=2, magnetic field, B1=N1μ0i2R1{B}_{1}=\dfrac{{N}_{1}{\mu }_{0}i}{2{R}_{1}}
For N2=5{N}_{2}=5
Radius of coil =R2=N1×R1N2={R}_{2}=\dfrac{{N}_{1}\times {R}_{1}}{{N}_{2}}
Magnetic field, B2=N2μ0i2R2{B}_{2}=\dfrac{{N}_{2}{\mu }_{0}i}{2{R}_{2}}
Thus, the ratio is B2B1=N2N1R1R2=N2N1×N2N1\dfrac{{B}_{2}}{{B}_{1}}=\dfrac{{N}_{2}}{{N}_{1}}\dfrac{{R}_{1}}{{R}_{2}}=\dfrac{{N}_{2}}{{N}_{1}}\times \dfrac{{N}_{2}}{{N}_{1}}
B2B1=254\dfrac{{B}_{2}}{{B}_{1}}=\dfrac{25}{4}

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