Question

The field intensity of two points on the axial
line of a bar magnet at distances 40cm and
20cm is 10 : 125. The length of the magnet is
1) 10 cm 2) 5 cm 3) 15 cm 4) 20 cm​

Answer 1

Answer:

i hope this is your answer .......

Answer 2

Answer:

A vector number that describes the state of any point that is magnetically influenced by a magnet, an electric current, or an electromagnetic wave, and is determined by the force that a free unit north pole applied at the site in question experiences when it is placed in a vacuum.

Explanation:

Given:

The field intensity of two points on the axial line of a bar magnet at distances of 40cm and 20cm is 10: 125

Find:

The length of the magnet
Solution:

                            $\begin{aligned}&\frac{\theta_{1}}{2}=\tan ^{-1}\left(\frac{1}{20}\right), \frac{\theta_{2}}{2}=\tan ^{-1}\left(\frac{1}{40}\right) \\&\mathrm{d}_{1}=\sqrt{\mathrm{l}^{2}+20^{2}} \\&\mathrm{~d}_{2}=\sqrt{\mathrm{l}^{2}+40^{2}} \\&\mathrm{~B}_{3}=\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{2}^{2}} \\&\mathrm{~B}_{4}=\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{2}^{2}} \\\end{aligned}$

                            $&\mathrm{~B}^{\prime \prime}=\sqrt{\mathrm{B}_{3}^{2}+\mathrm{B}_{4}^{2}+2 \mathrm{~B}_{3} \mathrm{~B}_{4}\left(\cos \pi-\theta_{2}\right)} \\\\&\cos \theta=\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2} \\\\&\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{1}^{2}} \\\\&\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{1}^{2}} \\\\&\mathrm{~B}_{2}=\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{1}^{2}}$

                             $\begin{aligned}&\mathrm{B}^{\prime}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}+2 \mathrm{~B}_{1} \mathrm{~B}_{2} \mathrm{COS}\left(\pi \theta_{1}\right)} \\&\frac{\mathrm{B}^{\prime}}{\mathrm{B}^{\prime \prime}}=\frac{10}{125}=\frac{\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{1}^{2}} \sqrt{2\left(1+\cos \theta_{1}\right)}}{\frac{\mu_{0} \mathrm{p}}{4 \pi \mathrm{d}_{2}^{2}} \sqrt{2\left(1-\cos \theta_{2}\right)}} \\\end{aligned}$

                                  $&\frac{2}{25}=\frac{\left(1^{2}+40^{2}\right) \sqrt{1-\left(\frac{1}{20} \mathrm{p}\right) /(1 / 20)^{2}}}{\left(1^{2}+20^{2}\right) \sqrt{1-\left(1-\left(\frac{1}{40}\right)^{2} /\left(1+4 \mathrm{~d}^{2}\right)\right.}} \\\\&1=20 \mathrm{~cm} .$

The field intensity of two points on the axial line of a bar magnet at distances of 40cm and 20cm is 10: 125. The length of the magnet is 20 cm.  That is  option (4)

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