Question

A 10 g bullet having a charge of 4.oo micro coulomb is fired at a speed of 270m/s in a horizontal direction. A vertical magnetic field of 500micro Tesla exist in the space . Find the deflection of the bullet due to the magnetic field as it travels through 100m . (1) 3.7 /10000 m (2) 3.7 /100000 m (3) 3.7/1000000 m (4)3.7/10000000 m

Answer 1

Answer on Question #74893, Physics / Electromagnetism

A 10 g bullet having a charge of 4.00 micro coulomb is fired at a speed of 270 m/s in a horizontal direction. A vertical magnetic field of 500 micro Tesla exist in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. (1) 3.7 / 10000 m (2) 3.7 / 100000 m (3) 3.7 / 1000000 m (4) 3.7 / 10000000 m.

Solution:

The Lorentz force law gives the magnitude of the force on the bullet at any time:

$$F = q v B$$ $$a = \frac {F}{m} = \frac {q v B}{m}$$

At the first instant, the acceleration of the bullet is (neglecting gravity) $a = 1.9 \times 10^{-8} \, \text{ms}^{-2}$

$$a = \frac {4 \times 10^{-6} C \times 270 \, \frac {m}{s} \times 500 \times 10^{-6} T}{10 \times 10^{-3} \, k g} = 5.410 \times 10^{-5} \, \text{ms}^2$$

As this acceleration is very small it leads to a negligible change in velocity so approximate the 100 m as being in the original direction of travel and the force as being constant, due to a constant velocity of 270 m/s. The time taken to cross 100 m is:

$$t = \frac {s}{v}$$ $$t = 100 / 270 \, s = 0.37 \, s$$

In this time, the sideways pushing magnetic force will lead to a small sideways deflection, $x$:

$$x = \frac {1}{2} a t ^ {2}$$ $$x = \frac {1}{2} \times 5.410 \times 10^{-5} \, \text{ms}^2 \times (0.37 \, s)^2 = 3.7 \times 10^{-6} \, m$$

Answer: $3.7 \times 10^{-6} \, m$

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