Question

A charged particle q enters a region of uniform magnetic field B (out of the page) and is deflected d after traveling a horizontal distance a. The magnitude of the momentum of the particle is–

• A. $\frac { q B } { 2 }$
• B. $\frac { \mathrm { qB } ^ { 0 } } { 2 }$
• C. Zero
• D. Not possible to be determined as it keeps changing

Answer 1

Answer: (A)

$\frac { q B } { 2 }$

Answer 2

Q AB == h(say)

In DOAD

Fig. cos q = … (i)

Now, In DACB sin $\left( \frac { \pi } { 2 } - \theta \right)$= … (ii) From (i) and (ii)

= \ h² = 2Rd R ==

Now, = qvB mv = qBR

Momentum = qB × $\left( \frac { a ^ { 2 } + d ^ { 2 } } { 2 d } \right)$