An electron moving with velocity $2 \times 10^7 \mathrm{~m} / \mathrm{s}$, describes a circle in a magnetic field of strength $2 \times {10^{ - 2}}T.$ If $\frac{\mathrm{e}}{\mathrm{m}}$ of electron is $1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$. Then the diameter of the circle will be

JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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An electron moving with velocity

2 \times 10^7 \mathrm{~m} / \mathrm{s}

, describes a circle in a magnetic field of strength

2 \times {10^{ - 2}}T.

\frac{\mathrm{e}}{\mathrm{m}}

of electron is

1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}

.
Then the diameter of the circle will be

View answer and explanation

Correct option: B

1.1 \mathrm{~cm}

Option B (

1.1 \mathrm{~cm}

) is correct because the diameter of the circular path is calculated using the given velocity, magnetic field, and charge-to-mass ratio.

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