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o-na [289]
3 years ago
7

A singly charged positive ion that has a mass of 6.07 ? 10-27 kg moves clockwise with a speed of 1.18 ? 104 m/s. the positively

charged ion moves in a circular path that has a radius of 2.75 cm. find the direction and strength of the uniform magnetic field through which the charge is moving. (hint: the magnetic force exerted on the positive ion is the centripetal force, and the speed given for the positive ion is its tangential speed.)
Physics
1 answer:
mezya [45]3 years ago
8 0
B = 0.018 T Ans, 

Since, it is moving in a circular path, thus, centripetal force will act on it i.e.
F =  \frac{ mv^{2} }{r}
where, m is the mass of the object, v is the velocity and r is the radius of circular path.

And, since a positive charge is moving, it will create magnetic force which is equal to F = qvB
where q is the charge, v is the velocity of the particle and B is the magnetic field.
Now, the two forces will be equal,
i.e. \frac{ mv^{2} }{r} = qvB
⇒ \frac{ mv }{r} = qB
⇒B = \frac{ mv }{qr}
<span>putting the values, we get,
</span>
use q = 1.6 * 10^ -19
⇒ B = 0.018 T

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\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

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\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

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\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

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\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

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