Question

A particle having a charge of 10 $\mu C$ and mass 1 $\mu g$ moves in a horizontal circle of radius 10 cm under the influence of a magnetic field of 0.1 T. When the particle is at a point P, a uniform electric field is switched on so that the particle starts moving along the tangent with uniform velocity. The electric field is

• A. 0.1 V/m
• B. 1.0 V/m
• C. 10.0 V/m
• D. 100 V/m

Answer 1

Answer: (C)

10.0 V/m

$\begin{array}{c}qVB=qE\\ Now,R=\frac{mv}{qB}\\ \Rightarrow V=\frac{BqR}{M}\\ \Rightarrow \frac{BqR}{M}\cdot B=E\\ \Rightarrow \frac{{\left(0.1\right)}^2\times \left(10\times {10}^{-6}\right)\times \left(10\times {10}^{-12}\right)}{\left(1\times {10}^{-6}\times {10}^{-3}\right)}=E\\ \Rightarrow \frac{{10}^{-2}\times 100\times {10}^{-2}}{{10}^{-3}}=E\\ \Rightarrow E=10V\end{array}$

hence,

option $\left(C\right)$ is correct answer.