CBSE Class 12 Physics 2024: question 25

Hints

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Step-by-step solution

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1. $\vec{B}$ is along $\hat{j}$. So the velocity component perpendicular to $\vec{B}$ is $v_x=1.0\times 10^7$ m/s, and the component parallel to $\vec{B}$ is $v_y=0.5\times 10^7$ m/s.

   Radius: $r=\frac{m{v}_x}{eB}$

   ½ mark

2. $r=\frac{9.1\times 10^{-31}\times 1.0\times 10^7}{1.6\times 10^{-19}\times 0.5\times 10^{-3}}=0.114\text{m}(\approx 11.4\text{cm})$

   ½ mark

3. Yes. The parallel component $v_y$ is unaffected by the field, so the electron also moves uniformly along $\vec{B}$. It therefore follows a helical path.

   ½ mark

4. Linear distance covered in one revolution (the pitch): $p=v_{y}T=v_y\frac{2\pi m}{eB}$

   ½ mark

5. $p=\frac{2\times 3.14\times 9.1\times 10^{-31}\times 0.5\times 10^7}{1.6\times 10^{-19}\times 0.5\times 10^{-3}}$

   ½ mark

6. $p\approx 0.357\text{m}\approx 0.36\text{m}$

   ½ mark

Answer: r ≈ 0.114 m; yes (helical path); linear distance per revolution ≈ 0.36 m

Revise the chapter

• Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
• Topic: Motion of a Charge in a Magnetic Field