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Question

Physics Question on Magnetic Field

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

A

250μT250\, \mu T

B

150μT150\, \mu T

C

125μT125\, \mu T

D

75μT75\, \mu T

Answer

250μT250\, \mu T

Explanation

Solution

Using formula B=μ0iR22(R2+X2)3/2B = \frac{\mu_{0}iR^{2}}{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\left(3\right)^{2}}{2\left[\left(3\right)^{2}+\left(4\right)^{2}\right]^{3/ 2}} At the centre of the coil, X=0X = 0 and B=μ0i2(3)B =\frac{\mu _{0}i}{2\left(3\right)} Using equation (i)\left(i\right) B=54×53(3)2×3B=250μT.B = \frac{54\times5^{3}}{\left(3\right)^{2}\times3} \Rightarrow B = 250\,\mu T.