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Question: The pole strength of \(12\;cm\) long bar magnet is having Magnetic moment \(20 \times 0.12\;A{m^2}\)...

The pole strength of 12  cm12\;cm long bar magnet is having Magnetic moment 20×0.12  Am220 \times 0.12\;A{m^2}. The magnetic induction at a point 10 cm10\,cm away from the center of the magnet on its axial line is [μo4π=10−7  Hm−1]\left[ {\dfrac{{{\mu _o}}}{{4\pi }} = {{10}^{ - 7}}\;H{m^{ - 1}}} \right]
(A) 1.17×10−3  T1.17 \times {10^{ - 3}}\;T
(B) 2.20×10−3  T2.20 \times {10^{ - 3}}\;T
(C) 1.17×10−2  T1.17 \times {10^{ - 2}}\;T
(D) 2.20×10−2  T2.20 \times {10^{ - 2}}\;T

Explanation

Solution

Use the information that the magnetic induction depends on the magnetic moment, pole strength, and bar magnet length. Information about these terms is given in the question. So will put given data directly in the magnetic induction expression for the determination of the magnitude of the magnetic induction.

Complete step by step solution:
It is given in the question that pole strength of the bar magnet is 20  Am20\;Am, length of the bar magnet is 2l, distance at which the magnetic induction is need to be determined from the center of the magnet is d=10  cmd = 10\;cm, magnitude of the magnetic moment is 20×0.12  Am220 \times 0.12\;A{m^2} and μo4π=10−7  Hm−1\dfrac{{{\mu _o}}}{{4\pi }} = {10^{ - 7}}\;H{m^{ - 1}}.
Write the expression of the magnetic induction on the axial line.

B=μo4π×2md(d2−l2)2B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2md}}{{{{\left( {{d^2} - {l^2}} \right)}^2}}}…. (1)

Here, mm is the magnetic moment, dd is the distance, and ll is the half of the bar magnet's length.

First, we will calculate the value of ll, so that we will use the above equation. We know that the length of the bar magnet is L=12  cmL = 12\;cm, therefore we get

l=L2l = \dfrac{L}{2}

Substitute the value of LL in the above equation.

l=12  cm2 l=6  cm\begin{array}{l} l = \dfrac{{12\;cm}}{2}\\\ l = 6\;cm \end{array}

We will now substitute the values in equation (1) to calculate magnetic induction.

Therefore, we get

B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2md}}{{{{\left( {{d^2} - {l^2}} \right)}^2}}}\\\ B = {10^{ - 7}}\;H{m^{ - 1}} \times \dfrac{{2\left( {20 \times 0.12\;A{m^2}} \right)\left( {10\;cm \times \dfrac{{{{10}^{ - 2}}\;m}}{{1\;cm}}} \right)}}{{{{\left( {{{\left( {10\;cm \times \dfrac{{{{10}^{ - 2}}\;m}}{{1\;cm}}} \right)}^2} - {{\left( {6\;cm \times \dfrac{{{{10}^{ - 2}}\;m}}{{1\;cm}}} \right)}^2}} \right)}^2}}}\\\ B = {10^{ - 7}}\;H{m^{ - 1}} \times 40 \times 0.12\;A{m^2} \times 2441\;{m^{ - 1}}\\\ B = 1.17 \times {10^{ - 3}}\;T \end{array}$$ Therefore, the magnetic induction at a point $10\,cm$ away from the center of the magnet on its axial line is $$1.17 \times {10^{ - 3}}\;T$$ and the option (A) is correct. **Note:** Remember the magnetic induction expression at some distance away from the center of the magnet and put the correct given values in magnetic induction expression because the incorrect value may result in the incorrect calculation and give the wrong answer.