
The distance of two points on the axis of a magnet from its centre are 10 cm and 20 cm respectively. The ratio of magnetic intensity at these points is 12.5 : 1. The length of the magnet will be
( a) 5 cm
(b) 25 cm
(c) 10 cm
( d) 20 cm
Answer
164.1k+ views
Hint:
In this question, we have given the distance of two points on the axis of magnet from its centre and ratio of magnetic intensity is also given. We have to find out the length of the magnet. First we use the formula of magnetic intensity at the axial point to solve this question . Now we put the values of r and B and by solving the equation, we get the length of the magnet.
Formula used:
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
Complete step by step solution:
We know the magnetic intensity on the axis of the magnet from its centre at a distance r is
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
The ratio of magnetic intensity on the axis at a distance ${{r}_{1}}$and ${{r}_{2}}$is ${{B}_{1}}$ and ${{B}_{2}}$
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{12.5}{1}$
Given ${{r}_{1}}$= 10 cm and ${{r}_{2}}$= 20 cm
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{u}_{0}}2{{r}_{1}}M}{4\pi {{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}}{\dfrac{{{u}_{0}}2{{r}_{2}}M}{4\pi {{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}}$
Which is equal to $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{r}_{1}}{{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}{{{r}_{2}}{{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}$
now we put the values in the above equation and we get
$\dfrac{12.5}{1}=\dfrac{10{{({{20}^{2}}-{{L}^{2}})}^{2}}}{20{{({{10}^{2}}-{{L}^{2}})}^{2}}}$
Hence by solving it, we get $\dfrac{5}{1}=\dfrac{{{20}^{2}}-{{L}^{2}}}{{{10}^{2}}-{{L}^{2}}}$
By cross multiplying, we get
$500-5{{L}^{2}}=400-{{L}^{2}}$
That is $4{{L}^{2}}=100$
Hence, we get L = 5 cm
We know length of the magnet is 2L
Then 2L = $2\times 5$ cm
L = 10 cm
Thus, Option (C ) is correct.
Note:
Students make mistakes while comparing the equations. We should take care while comparing the equations to get the correct answer. Also recall the complete formula of axial and equatorial magnetic field as they can be used in such type of questions.
In this question, we have given the distance of two points on the axis of magnet from its centre and ratio of magnetic intensity is also given. We have to find out the length of the magnet. First we use the formula of magnetic intensity at the axial point to solve this question . Now we put the values of r and B and by solving the equation, we get the length of the magnet.
Formula used:
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
Complete step by step solution:
We know the magnetic intensity on the axis of the magnet from its centre at a distance r is
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
The ratio of magnetic intensity on the axis at a distance ${{r}_{1}}$and ${{r}_{2}}$is ${{B}_{1}}$ and ${{B}_{2}}$
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{12.5}{1}$
Given ${{r}_{1}}$= 10 cm and ${{r}_{2}}$= 20 cm
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{u}_{0}}2{{r}_{1}}M}{4\pi {{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}}{\dfrac{{{u}_{0}}2{{r}_{2}}M}{4\pi {{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}}$
Which is equal to $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{r}_{1}}{{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}{{{r}_{2}}{{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}$
now we put the values in the above equation and we get
$\dfrac{12.5}{1}=\dfrac{10{{({{20}^{2}}-{{L}^{2}})}^{2}}}{20{{({{10}^{2}}-{{L}^{2}})}^{2}}}$
Hence by solving it, we get $\dfrac{5}{1}=\dfrac{{{20}^{2}}-{{L}^{2}}}{{{10}^{2}}-{{L}^{2}}}$
By cross multiplying, we get
$500-5{{L}^{2}}=400-{{L}^{2}}$
That is $4{{L}^{2}}=100$
Hence, we get L = 5 cm
We know length of the magnet is 2L
Then 2L = $2\times 5$ cm
L = 10 cm
Thus, Option (C ) is correct.
Note:
Students make mistakes while comparing the equations. We should take care while comparing the equations to get the correct answer. Also recall the complete formula of axial and equatorial magnetic field as they can be used in such type of questions.
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