Question

A beam of protons enters a uniform magnetic field of 0.3T with a velocity of $4\times {10}^{5}m/s$ in a direction making an angle of ${60}^0$ with the direction of magnetic field. The path of motion of the particle will be

• A. circular
• B. straight line
• C. spiral
• D. helical

Answer 1

Answer: (D)

helical

Given: $B=0.3T$

$v=4\ast {10}^{5}m/s$

$\theta ={60}^{\circ }$ //angle of proton beam with the $di{r}^n$of magnetic field//

To find: path of motion determination$=?$

Solution: Now, we have to find out

the component of proton's velocity,

$v_{||}=vcos{60}^{\circ }$

$=>v_{||}=4\ast {10}^5\ast 0.5==>v_{||}=2\ast {10}^{5}m/s$

$v_{\perp }=vsin{60}^{\circ }$

$=>v_{\perp }=4\ast {10}^5\ast 0.866==>v_{\perp }=3.464\ast {10}^{5}m/s$

now, it is clear that,

non-zero $v_{||}$ makes the proton to move along the field direction.

and

non-zero $v_{\perp }$ makes the proton to move along a circular path.

hence, path of motion is helical.

The correct opt: D