A charged particle enters a region of magnetic field at an angle - and then leaves it at an angle - as shown in figure- If magnetic field strength increases along x-axis- then the correct relationship between - and - isC- -D- None

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Standard XII

Physics

Time Period Independent of Amplitude

A charged par...

Question

A charged particle enters a region of magnetic field at an angle $α$ and then leaves it at an angle $β$ as shown in figure. If magnetic field strength increases along x-axis, then the correct relationship between $α$ and $β$ is

α > β

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α = β

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α < β

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None

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Solution

The correct option is C $α < β$
Radius $r = \frac{m v}{q B}$ . Value of B is more at exit, so the radius is lesser at the exit. So, the path is not symmetrical and the deviation is more in the second half of the field than the first half. Therefore, $β > α$ .

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A charged particle enters a region of magnetic field at an angle α and then leaves it at an angle β as shown in figure. If magnetic field strength increases along x-axis, then the correct relationship between α and β is

The correct option is C α<βRadius r=mvqB. Value of B is more at exit, so the radius is lesser at the exit. So, the path is not symmetrical and the deviation is more in the second half of the field than the first half. Therefore, β>α.

A charged particle enters a region of magnetic field at an angle $α$ and then leaves it at an angle $β$ as shown in figure. If magnetic field strength increases along x-axis, then the correct relationship between $α$ and $β$ is

The correct option is C $α < β$
Radius $r = \frac{m v}{q B}$ . Value of B is more at exit, so the radius is lesser at the exit. So, the path is not symmetrical and the deviation is more in the second half of the field than the first half. Therefore, $β > α$ .