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Standard XII

Physics

More about Magnetic Dipoles

The pitch of ...

Question

The pitch of the helical path of the motion of the particle will be

\frac{π}{100} m

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\frac{π}{125} m

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\frac{π}{215} m

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\frac{π}{250} m

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Solution

The correct option is D $\frac{π}{250} m$
$Given:-$ $(\frac{q}{m})$ = $\frac{1063}{19}$ C/Kg

$\to B = (20^i + 30^j + 50^k)$ T

$\to v = (20^i + 50^j + 30^k)$ m/s

$To find:-$ pitch of the helical path

$Solution:-$

Pitch= TV cos $θ$

= $\frac{1}{f} V \frac{\to V \to B}{B V} = \frac{π \sqrt{38} \times 400}{10^{4} \sqrt{3800}} = \frac{4 π}{10^{3}} m = \frac{π}{250} m$

$Hence the correct option is D$

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The pitch of the helical path of the motion of the particle will be

The correct option is D π250mGiven:- (qm)= 106319 C/Kg →B=(20^i+30^j+50^k)T →v=(20^i+50^j+30^k)m/sTo find:-pitch of the helical pathSolution:-Pitch= TV cos θ =1fV→V→BBV=π√38×400104√3800=4π103m=π250mHence the correct option is D