An electron moves at right angle to a magnetic field of 1.5 × 10^ -2 ~T with a speed of 6 × 10^7 ~m / s . If the specific charge of the electron is 1.7 × 10^ 11 C / kg . The radius of the circular path will be

Correct answer: C. 2.35 cm The formula for radius of circular path is r=(m v)/(e B)=(v)/(((e)/(m)) B) ...(1) Given : e / m of electron aligned & =1.7 × 10^ 11 C / kg and v=6 × 10^7 m / s B & =1.5 × 10^ -2 T aligned aligned So, r & = 6 × 10^7 1.7 × 10^ 11 × 1.5 × 10^ -2 & =2.35 × 10^ -2 m=2.35 cm aligned

JIPMER 2005

Exam: JIPMER
Year: 2005
Subject: Physics
Chapter: Magnetic Effects of Current

JIPMER 2005 · Physics · Magnetic Effects of Current

An electron moves at right angle to a magnetic field of 1.5 × 10^...

An electron moves at right angle to a magnetic field of

1.5 \times 10^{-2} \mathrm{~T}

with a speed of

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

. If the specific charge of the electron is

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

. The radius of the circular path will be

Options

A2.9 cm
B3.9 cm
C2.35 cm
D2 cm

Correct answer

C. 2.35 cm

Step-by-step solution

The formula for radius of circular path is

r=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}

...(1) Given

: e / m

of electron

\begin{aligned} & =1.7 \times 10^{11} \mathrm{C} / \mathrm{kg} \text{ and } v=6 \times 10^7 \mathrm{ m} / \mathrm{s} \\ B & =1.5 \times 10^{-2} \mathrm{ T} \end{aligned}

\quad

\begin{aligned} So, r & =\frac{6 \times 10^7}{1.7 \times 10^{11} \times 1.5 \times 10^{-2}} \\ & =2.35 \times 10^{-2} \mathrm{ m}=2.35 \mathrm{ cm}\end{aligned}

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