Question

If a particle of charge ${10}^{-12}C$ moving along the x-axis with a velocity ${10}^5$ m/s. experiences a force of ${10}^{-10}N$ in y-direction due to magnetic field, then the minimum magnetic field is.

• A. $6.25\times {10}^{3}T$ in Z direction
• B. ${10}^{-15}T$ in Z direction
• C. $6.25\times {10}^{-3}T$ in Z direction
• D. ${10}^{-3}T$ in -ve Z direction

Answer 1

Answer: (D)

${10}^{-3}T$ in -ve Z direction

$f=qVB\sin \theta$
B will be minimum when $\theta ={90}^{\circ }$
$B=\frac{f}{qV}$
=$\frac{{10}^{-10}}{{10}^{-12}\times {10}^5}$
=${10}^{-3}T$ in Z-direction