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Question: A circular loop of radius 3 cm is having a current of 12.5 A. The magnitude of magnetic field at a d...

A circular loop of radius 3 cm is having a current of 12.5 A. The magnitude of magnetic field at a distance of 4 cm on its axis is
A. 5.65×10−5T5.65\times {{10}^{-5}}T
B. 5.27×10−5T5.27\times {{10}^{-5}}T
C. 6.54×10−5T6.54\times {{10}^{-5}}T
D. 9.20×10−5T9.20\times {{10}^{-5}}T

Explanation

Solution

To solve the given question, we must know about the magnetic field produced by a current carrying loop. Use the formula for the magnitude of the magnetic field at a point on the axis of the loop at a distance x from its centre.

Formula used:
B=μ0iR22(R2+x2)3/2B=\dfrac{{{\mu }_{0}}i{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}

Complete step by step answer:
When a current flows through a conducting wire, it produces a magnetic field around it.
In this case, the conducting wire is a form of a loop. The magnitude of the magnetic field at a point on the axis of the loop at a distance x from its centre is given as B=μ0iR22(R2+x2)3/2B=\dfrac{{{\mu }_{0}}i{{R}^{2}}}{2{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}} …. (i),
where R is the radius of the loop, i is the current flowing in the loop and μ0{{\mu }_{0}} is permeability of free space (It is a constant).
It is given that the radius of the loop is 3cm. Therefore, R=3cm=3×10−2mR=3cm=3\times {{10}^{-2}}m.
The current flowing in the given loop is 12.5A. This means that i=12.5Ai=12.5A.
And x=4cm=4×10−2mx=4cm=4\times {{10}^{-2}}m.
The value of permeability of free space is μ0=4π×10−7TmA−1{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}.
Substitute the values of i, R, x and μ0{{\mu }_{0}} in equation (i).
⇒B=(4π×10−7)(12.5)(3×10−2)22((3×10−2)2+(4×10−2)2)3/2=5.65×10−5T\Rightarrow B=\dfrac{\left( 4\pi \times {{10}^{-7}} \right)\left( 12.5 \right){{\left( 3\times {{10}^{-2}} \right)}^{2}}}{2{{\left( {{\left( 3\times {{10}^{-2}} \right)}^{2}}+{{\left( 4\times {{10}^{-2}} \right)}^{2}} \right)}^{3/2}}}=5.65\times {{10}^{-5}}T
This means that the magnitude of the magnetic field at a point on its axis, at distance of 4cm from its centre of the given current carrying loop is equal to 5.65×10−5T5.65\times {{10}^{-5}}T.

So, the correct answer is “Option A”.

Note:
In all the four given options, we can see that the unit of magnetic field is tesla (T), which is an SI unit of magnetic field.
Therefore, we have to substitute the values of the quantities in their SI units. Hence, we substituted x and R in metres (m).
The direction of the magnetic field is given by the right thumb rule.