A beam of ions with a velocity of 2- 105m-s- enters normally into a uniform magnetic field of 4- 10-2T -If the specific charge of the ion is 5- 107 C-kg- the radius of the circular path described is -

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Byju's Answer

Standard XII

Physics

Circular Kinematics

A beam of ion...

Question

A beam of ions with a velocity of $2 \times 10^{5} m / s$ . enters normally into a uniform magnetic field of $4 \times 10^{- 2} T$ .If the specific charge of the ion is $5 \times 10^{7} C / k g$ , the radius of the circular path described is :

0.10 m

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0.16 m

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0.20 m

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0.25 m

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Solution

The correct option is A $0.10 m$
Radius of the charged particle in magnetic field is,
$r = \frac{m v}{q B}$
= $\frac{2 \times 10^{5}}{4 \times 10^{- 2} \times 5 \times 10^{- 7}} m$
$= 0.10 m$

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A beam of ions with a velocity of 2×105m/s. enters normally into a uniform magnetic field of 4×10−2T .If the specific charge of the ion is 5×107C/kg, the radius of the circular path described is :

The correct option is A 0.10mRadius of the charged particle in magnetic field is,r=mvqB=2×1054×10−2×5×10−7m=0.10m

A beam of ions with a velocity of $2 \times 10^{5} m / s$ . enters normally into a uniform magnetic field of $4 \times 10^{- 2} T$ .If the specific charge of the ion is $5 \times 10^{7} C / k g$ , the radius of the circular path described is :

The correct option is A $0.10 m$
Radius of the charged particle in magnetic field is,
$r = \frac{m v}{q B}$
= $\frac{2 \times 10^{5}}{4 \times 10^{- 2} \times 5 \times 10^{- 7}} m$
$= 0.10 m$