Magnetic Effects & Magnetism

JEE Main level, single correct, medium

56 of 276

Question 56 of 276

The magnetic field at the center of current carrying circular loop is B1{B}_{1}. The magnetic field at a distance of 3\sqrt{3} times radius of the given circular loop from the center on its axis is B2{B}_{2}. The value of B1B2\dfrac{{B}_{1}}{{B}_{2}} will be

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Answer: (C) 8:18:1

Magnetic field at the centre of the loop i.e. B1=μ0i2R{B}_{1}=\dfrac{{\mu }_{0}i}{2R}
Magnetic field on the axis of the loop is,
B2=μ0iR22(R2+x2)32=μ0i×R22(R2+3R2)32{B}_{2}=\dfrac{{\mu }_{0}i{R}^{2}}{2{\left({R}^{2}+{x}^{2}\right)}^{\dfrac{3}{2}}}=\dfrac{{\mu }_{0}i\times {R}^{2}}{2{\left({R}^{2}+3{R}^{2}\right)}^{\dfrac{3}{2}}}=μ0i×R22×8R3=\dfrac{{\mu }_{0}i\times {R}^{2}}{2\times 8{R}^{3}}=μ0i2×8R=\dfrac{{\mu }_{0}i}{2\times 8R}
Therefore, the ratio B1B2=81\dfrac{{B}_{1}}{{B}_{2}}=\dfrac{8}{1}

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