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Byju's Answer

Standard XII

Physics

Circular Kinematics

Magnetic fiel...

Question

Magnetic field of magnitude $B_{0}$ exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

\frac{2 π m θ}{q B π}

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\frac{2 π m}{q B} (1 - \frac{θ}{π})

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\frac{m θ}{q B}

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\frac{2 m θ}{q B} (\frac{1}{θ} - \frac{1}{π})

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Solution

The correct option is B $\frac{2 π m}{q B} (1 - \frac{θ}{π})$
By geometry, $α = 2 θ$
Here, $T = \frac{2 π R}{V} = \frac{2 π m}{q B}$
So time $t = \frac{T}{2 π} \times (2 π - α)$
$t = T (1 - \frac{α}{2 π}) = T (1 - \frac{θ}{π})$
$= \frac{2 π m}{q B} (1 - \frac{θ}{π})$

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Magnetic field of magnitude B0 exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

The correct option is B 2πmqB(1−θπ)By geometry, α=2θ Here, T=2πRV=2πmqB So time t=T2π×(2π−α) t=T(1−α2π)=T(1−θπ) =2πmqB(1−θπ)

Magnetic field of magnitude $B_{0}$ exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

The correct option is B $\frac{2 π m}{q B} (1 - \frac{θ}{π})$
By geometry, $α = 2 θ$
Here, $T = \frac{2 π R}{V} = \frac{2 π m}{q B}$
So time $t = \frac{T}{2 π} \times (2 π - α)$
$t = T (1 - \frac{α}{2 π}) = T (1 - \frac{θ}{π})$
$= \frac{2 π m}{q B} (1 - \frac{θ}{π})$