Question

Consider a bar magnet having magnetic length 10cm and pole strength 7.5Am. What will be the magnitude of magnetic field at a distance of 10 cm from the center of the magnet on the line, which is along the axis of magnet?​

Answer 1

Given:

A bar magnet having magnetic length 10cm and pole strength 7.5 Am.

To find:

Magnitude of magnetic field at a distance of 10 cm from magnet.

Calculation:

General expression for magnetic field intensity on the axis of a magnetic dipole is given as:

$B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{2Mx}{ {( {x}^{2} - {l}^{2} )}^{2} } \bigg \}$

Putting available values in SI unit:

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{2 \times 7.5 \times ( \frac{10}{100}) }{ {( {0.1}^{2} - {0.05}^{2} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{15 \times ( \frac{10}{100}) }{ {( {0.1}^{2} - {0.05}^{2} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{15 \times 0.1}{ {( {0.1}^{2} - {0.05}^{2} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{15 \times 0.1}{ {( {10}^{ - 2} - 0.25 \times {10}^{ - 2} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{15 \times 0.1}{ {( 0.75 \times {10}^{ - 2} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{15 \times 0.1}{ {( 75 \times {10}^{ - 4} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{1.5 }{ {( 75 \times {10}^{ - 4} )}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{1.5 }{ (5625 \times {10}^{ - 8} )} \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{1.5 \times {10}^{8} }{ 5625 } \bigg \}$

$= > B = {10}^{ - 7} \times \bigg \{ \dfrac{1.5 \times {10}^{8} }{ 5625 } \bigg \}$

$= > B = \dfrac{1.5 \times {10}^{1} }{ 5625 }$

$= > B = \dfrac{15 }{ 5625 }$

$= > B = 26.6 \times {10}^{ - 4} \: tesla$

So, final answer is:

$\boxed{ \bf{ B = 26.6 \times {10}^{ - 4} \: tesla}}$