Question

An electron at point A in the figure has a speed $V_0$ of magnitude $1.41\times {10}^{6}m/s$. It enters into a uniform magnetic field and follows a semicircular path in it as shown.


The magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B is?

• A. $1.60\times {10}^{-4}T$
• B. $1.60\times {10}^{-2}T$
• C. $1.6T$
• D. $1.60\times {10}^{-3}T$

Answer 1

The correct option is A $1.60\times {10}^{-4}T$

At point A, magnetic force should be towards right. Thus fleming's left hand rule suggests that magnetic field should be inside the plane of the paper.
The electron experiences centripetal force under the action of magnetic field.
Hence $\frac{m{v}^2}{r}=qvB$
$\Rightarrow B=\frac{mv}{qr}$
$=1.6\times {10}^{-4}T$