A particle of charge per unit mass - is released from origin with velocity v - v 0- in a magnetic field B - B 0k-for x -3-2 v 0- B 0-And B -0 for x -3-2 v 0- B 0-The x co-ordinate of the particle at time t-3 B0-A- -3-2 V 0- B 0- V 0 t -3 B 0-B- -3-2 V 0- B 0- V 01-2 t -3 B 0-C- -3-2 V 0- B 0-3-2 V 0 t - B 0-D- -3-2 V 0- B 0- V 01-2 t

The correct option is B √(3)/2V_0/B_0α+V_01/2 ( t π/3B_0α )Angle turned by R in the field is 60^∘ Time taken to cross field is π/3B_0α Additional distance = ...

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Standard XII

Physics

The Circular Motion

A particle of...

Question

A particle of charge per unit mass $α$ is released from origin with velocity $\to v = v_{0}^i$ in a magnetic field $\to B = - B_{0}^k$
for $x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α}$ ,
And $\to B = 0 f o r x > \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α}$ ,
The x co-ordinate of the particle at time $t (> \frac{π}{3 B_{0} α})$

\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + \frac{\sqrt{3}}{2} V_{0} (t - \frac{π}{B_{0} α})

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\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + V_{0} (t - \frac{π}{3 B_{0} α})

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\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + V_{0} \frac{1}{2} (t - \frac{π}{3 B_{0} α})

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\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + V_{0} \frac{1}{2} t

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Solution

The correct option is C $\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + V_{0} \frac{1}{2} (t - \frac{π}{3 B_{0} α})$
Angle turned by R in the field is $60^{\circ}$
Time taken to cross field is $\frac{π}{3 B_{0} α}$
Additional distance = $v_{0} c o s θ (t - \frac{π}{3 B_{0} α})$

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A particle of charge per unit mass α is released from origin with velocity →v=v0^i in a magnetic field →B=−B0^k for x≤√32v0B0α, And →B=0 for x>√32v0B0α, The x co-ordinate of the particle at time t(>π3B0α)

The correct option is C √32V0B0α+V012(t−π3B0α)Angle turned by R in the field is 60∘ Time taken to cross field is π3B0α Additional distance = v0cosθ(t−π3B0α)

The correct option is C $\frac{\sqrt{3}}{2} \frac{V_{0}}{B_{0} α} + V_{0} \frac{1}{2} (t - \frac{π}{3 B_{0} α})$
Angle turned by R in the field is $60^{\circ}$
Time taken to cross field is $\frac{π}{3 B_{0} α}$
Additional distance = $v_{0} c o s θ (t - \frac{π}{3 B_{0} α})$