The force between two magnetic poles is $ F $ . if the distance between the poles and pole strength of each pole is doubled, then the force experienced is ............. :
Answer
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Hint: The force between two magnetic poles is proportional to the product of the strength of the magnetic poles and inversely proportional to the square of the distance between these two magnetic poles. We will use this relation to determine the force experienced when the pole strength is doubled and the distance between the poles is doubled.
Formula used
In this question, we will use this formula
$ \Rightarrow F = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
where, $ F $ is the force between two magnetic poles of strength $ {M_1} $ and $ {M_2} $ at a distance of $ d $ .
Complete Step-by-step solution
We know that the force between two magnetic poles of strength $ {M_1} $ and $ {M_2} $ at a distance $ d $ can be calculated as:
$ \Rightarrow F = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
Now, we’ve been stated that the distance between the two poles is doubled i.e. $ d' = 2d $ and the strength of each pole is doubled, which suggests $ {M_1}^\prime = 2{M_1} $ and $ {M_2}^\prime = 2{M_2} $ .
Thus, the new force between these two poles will be,
$ F' = \dfrac{{{\mu _0}{M_1}^\prime {M_2}^\prime }}{{4\pi {{d'}^2}}} $
On substituting $ d' = 2d $ , $ {M_1}^\prime = 2{M_1} $ and $ {M_2}^\prime = 2{M_2} $ , we get,
$ \Rightarrow F' = \dfrac{{{\mu _0}\left( {2{M_1}} \right)\left( {2{M_2}} \right)}}{{4\pi {{\left( {2d} \right)}^2}}} $
$ \Rightarrow F' = \dfrac{4}{4}\dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
On further solving the equation we get,
$ \Rightarrow F' = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
$ \Rightarrow F' = F $
Hence the new force between the magnetic poles if the distance between them gets doubled and their pole strength also gets doubled remains unchanged i.e., the force between them is $ F $ .
Note
Here we have assumed that the poles are in a vacuum which means there is no medium between them and there are no other magnetic poles in their vicinity that will affect the force they experience. We must be careful since we’ve asked to find the new force between the poles as compared to the old scenario. Since the force remains unchanged in the new scenario.
Formula used
In this question, we will use this formula
$ \Rightarrow F = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
where, $ F $ is the force between two magnetic poles of strength $ {M_1} $ and $ {M_2} $ at a distance of $ d $ .
Complete Step-by-step solution
We know that the force between two magnetic poles of strength $ {M_1} $ and $ {M_2} $ at a distance $ d $ can be calculated as:
$ \Rightarrow F = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
Now, we’ve been stated that the distance between the two poles is doubled i.e. $ d' = 2d $ and the strength of each pole is doubled, which suggests $ {M_1}^\prime = 2{M_1} $ and $ {M_2}^\prime = 2{M_2} $ .
Thus, the new force between these two poles will be,
$ F' = \dfrac{{{\mu _0}{M_1}^\prime {M_2}^\prime }}{{4\pi {{d'}^2}}} $
On substituting $ d' = 2d $ , $ {M_1}^\prime = 2{M_1} $ and $ {M_2}^\prime = 2{M_2} $ , we get,
$ \Rightarrow F' = \dfrac{{{\mu _0}\left( {2{M_1}} \right)\left( {2{M_2}} \right)}}{{4\pi {{\left( {2d} \right)}^2}}} $
$ \Rightarrow F' = \dfrac{4}{4}\dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
On further solving the equation we get,
$ \Rightarrow F' = \dfrac{{{\mu _0}{M_1}{M_2}}}{{4\pi {d^2}}} $
$ \Rightarrow F' = F $
Hence the new force between the magnetic poles if the distance between them gets doubled and their pole strength also gets doubled remains unchanged i.e., the force between them is $ F $ .
Note
Here we have assumed that the poles are in a vacuum which means there is no medium between them and there are no other magnetic poles in their vicinity that will affect the force they experience. We must be careful since we’ve asked to find the new force between the poles as compared to the old scenario. Since the force remains unchanged in the new scenario.
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