A 10 g bullet having a charge of 4-00 - C is fired at a speed of 270 m s -1 in a horizontal direction- A vertical magnetic field of 500 - T exists in the space- Find the deflection of the bullet due to the magnetic field as it travels through 100 m- Make appropriate approximations-

We know, F = q (V × B) m α = q VB α = q VB/m Time taken to travel 100 m distance in horizontal i = S/v = 100/270 sec. So, in this time intervel bullet deflectio ...

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Standard XII

Physics

The Circular Motion

A 10 g bullet...

Question

A 10 g bullet having a charge of 4.00 $μ C$ is fired at a speed of 270 m $s^{- 1}$ in a horizontal direction. A vertical magnetic field of 500 $μ T$ exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.

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Solution

We know,

$F = q (V \times B)$

m $α$ = q VB

$α = \frac{q V B}{m}$

Time taken to travel 100 m distance in horizontal

$i = \frac{S}{v} = \frac{100}{270}$ sec.

So, in this time intervel bullet deflection is

$y = \frac{1}{2} a t^{2}$

$= \frac{1}{2} \times \frac{4.00 \times 10^{- 6} \times 270 \times 500 \times 10^{- 6}}{10 \times 10^{- 3}} \times {(\frac{100}{270})}^{2}$

$= 3.7 \times 10^{- 6}$ m.

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A 10 g bullet having a charge of 4.00 μC is fired at a speed of 270 m s−1 in a horizontal direction. A vertical magnetic field of 500 μT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.

We know, F=q(V×B) m α = q VB α=qVBm Time taken to travel 100 m distance in horizontal i=Sv=100270 sec. So, in this time intervel bullet deflection is y=12 at2 =12×4.00×10−6×270×500×10−610×10−3×(100270)2 =3.7×10−6 m.

A 10 g bullet having a charge of 4.00 $μ C$ is fired at a speed of 270 m $s^{- 1}$ in a horizontal direction. A vertical magnetic field of 500 $μ T$ exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.