Magnetic Effects & Magnetism

JEE Main level, single correct, medium

102 of 276

Question 102 of 276

The magnetic field due to a current carrying circular loop of radius 3 cm3 \mathrm{~cm} at a point on the axis at a distance of 4 cm4 \mathrm{~cm} from the centre is 54μT54 \mu \mathrm{T}. What will be its value at the centre of the Loop?

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Answer: (A) 250μT250 \mu \mathrm{T}

Using formula B=μ0II 2(R2+X2)3/2B=\frac{\mu_0^{\text {II }}}{2\left(R^2+X^2\right)^{3 / 2}}, we get
54=μ0i(3)22[(3)2+(4)2]3/254=\frac{\mu_0 i(3)^2}{2\left[(3)^2+(4)^2\right]^{3 / 2}}
At the centre of the coil, X=0X=0 and B=μ0i2(3)B=\frac{\mu_0 i}{2(3)}
Using equation (i)
B=54×53(3)2×3⇒B=250μT. B=\frac{54 \times 5^3}{(3)^2 \times 3} \Rightarrow B=250 \mu \mathrm{T} \text {. }

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