Question

In Thomson’s experiment, a magnetic field of induction ${10}^{-2}$ $wb/m^2$ is used. For an undeflected beam of cathode rays, a Potential Difference of $500V$ is applied between the plates which are $0.5cm$ apart. Then the velocity of the cathode ray beam is _____ m/s.

• A. $4\times {10}^7$
• B. $2\times {10}^7$
• C. $2\times {10}^8$
• D. ${10}^7$

Answer 1

The correct option is D ${10}^7$

$B={10}^{-2}Wb/m^2$
So, from Potential difference and distance, $E=\frac{500\times {10}^3}{0.5}={10}^5$ V/m
So, for Magnetic field force $=qVB$------------------ (I)
The force due to electric field $=qE$---------------(II)
Equating (I) and (II) acting in opposite directions,
$qVB=qE$
$\Rightarrow {10}^5={10}^{-2}\times V\Rightarrow V={10}^{7}m/s$