A 5 mg particle carrying a charge of is moving with velocity of in a region having magnetic field . It moves a

Question

A 5 mg particle carrying a charge of $5 π \times 10^{- 6} C$ is moving with velocity of $(3 \hat{i} + 2 \hat{k}) \times 10^{- 2} m/s$ in a region having magnetic field $\vec{B} = 0.1 \hat{k} {Wb/m}^{2}$ . It moves a distance of $α$ meter along $\hat{k}$ when it completes 5 revolutions. The value of $α$ is ________. [2026]

Smart Achievers · Accepted Answer

(2)

$T = \frac{2 π m}{q B}$

$Pitch p = V T = v (\frac{2 π m}{q B})$

$α = 5 \times p = 5 \times v_{∥} (\frac{2 π m}{q B})$

$\Rightarrow α = 5 \times 0.02 \times \frac{2 π \times 5 \times 10^{- 6}}{5 π \times 10^{- 6} \times 0.1} = 2 m$

Solution

Ans.
(2)

$T = \frac{2 π m}{q B}$

$Pitch p = V T = v (\frac{2 π m}{q B})$

$α = 5 \times p = 5 \times v_{∥} (\frac{2 π m}{q B})$

$\Rightarrow α = 5 \times 0.02 \times \frac{2 π \times 5 \times 10^{- 6}}{5 π \times 10^{- 6} \times 0.1} = 2 m$

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Solution

Q. A 5 mg particle carrying a charge of 5π×10-6 C is moving with velocity of (3i^+2k^)×10-2 m/s in a region having magnetic field B→=0.1k^ Wb/m2. It moves a distance of α meter along k^ when it completes 5 revolutions. The value of α is ________. [2026]

Ans. (2) T=2πmqB Pitch p=VT=v(2πmqB) α=5×p=5×v∥(2πmqB) ⇒α=5×0.02×2π×5×10-65π×10-6×0.1=2 m

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Ans.
(2)

$T = \frac{2 π m}{q B}$

$Pitch p = V T = v (\frac{2 π m}{q B})$

$α = 5 \times p = 5 \times v_{∥} (\frac{2 π m}{q B})$

$\Rightarrow α = 5 \times 0.02 \times \frac{2 π \times 5 \times 10^{- 6}}{5 π \times 10^{- 6} \times 0.1} = 2 m$