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Standard X

Physics

Magnetic Field Due to Current in a Solenoid

A uniform ele...

Question

A uniform electric field E is present horizontally along the paper throughout region but uniform magnetic field B0 is present horizontally (perpendicular to plane of paper in inward direction) right to the line AB as shown. A charge particle having charge q and mass m is projected vertically upward and crosses the line AB after time t0. Find the speed of projection if particle moves after t0 with constant velocity. (given qE = mg)

g t_{0}

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2 g t_{0}

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\frac{g t_{0}}{2}

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particle can’t move with constant velocity after crossing AB

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Solution

The correct option is B $2 g t_{0}$
When crosses AB
$g v B_{0} c o s θ = m g$
$g v B_{0} s i n θ = q E$
$t a n θ = \frac{q E}{m g} = 1$
$θ = \frac{π}{4}$
along horizontal $v c o s θ = \frac{q E}{m} t_{0}$
$u - g t_{0} = v s i n θ$
$u = (g + \frac{q E}{m}) t_{0} = 2 g t_{0}$

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The correct option is B 2gt0When crosses AB gvB0 cos θ=mg gvB0 sin θ=qE tan θ=qEmg=1 θ=π4 along horizontal v cos θ=qEmt0 u−gt0=v sin θ u=(g+qEm)t0=2gt0

The correct option is B $2 g t_{0}$
When crosses AB
$g v B_{0} c o s θ = m g$
$g v B_{0} s i n θ = q E$
$t a n θ = \frac{q E}{m g} = 1$
$θ = \frac{π}{4}$
along horizontal $v c o s θ = \frac{q E}{m} t_{0}$
$u - g t_{0} = v s i n θ$
$u = (g + \frac{q E}{m}) t_{0} = 2 g t_{0}$