Question

An electron has a speed $\sqrt{2}\times {10}^{6}m{s}^{-1}$ at $A$ as shown in Fig. Find the direction and magnitude of magnetic field so that electron reaches $B$ following a semicircular path.

• A. $1.6\times {10}^{-4}T\otimes$
• B. $1.6\times {10}^{-4}T\sqcap$
• C. $3.6\times {10}^{-4}T\otimes$
• D. none of these

Answer 1

The correct option is A $1.6\times {10}^{-4}T\otimes$

As seen from the figure diameter of the semi-circular track is 10 cm.
$r=\frac{d}{2}=5cm$
$r=\frac{mv}{qB}$
$\therefore B=\frac{mv}{qr}$
$\Rightarrow B=\frac{9\times {10}^{-31}\times \sqrt{2}\times {10}^6}{1.6\times {10}^{-19}\times 5\times {10}^{-2}}=\frac{9\sqrt{2}}{8}{10}^{-4}=1.6\times {10}^{-4}T$