# Deflection of a object due to magnetic field

1. Jan 5, 2005

### joej

an object of 3.8g moves @ 180m/s perpendicular to a magnetic field of (5x10^-5)T

the object possesses a net charge of (8.1x10^-9)C

by what distance will it be deflected from its pathe due to the magnetic field after it has travelled 1.0 km

so I'm guessing I'll have to find out the radius of the curvature here, so...

F = ma

a = v^2/r

qvB = (mv^2)/r

r = (mv)/(qB)

r = (1.69x10^12)m

and..... yeah.... where do I go from there.... or do I have to do something different?

2. Jan 5, 2005

### dextercioby

Are u sure this is the original text of the problem,with these figures??I mean the curvature radius is larger than the mean distance Earth-Sun,the deflectio will be almost zero.

Daniel.

3. Jan 5, 2005

### joej

yeah I'm pretty sure... just double check it now again...

m = 3.8 g
B = 5 x 10 ^ -5 T
q = 8.1 x 10 ^ -9 C
distance travelled = 1.0 km

4. Jan 5, 2005

### dextercioby

I'll let u do the calculations numerically,i'll simply provide the formulas.call the distance of 1Km by d,the curvature radius by "R" and the angle subtented by 'd' $\theta$
$$d=R\theta$$ (1)
$$deflection=R-R\cos\theta$$ (2)

Since the angle is vanishingly small,u may use that,for very,very small arguments
$$\cos\theta\sim 1-\frac{\theta^{2}}{2}$$ (3)

Daniel.

5. Jan 5, 2005

### joej

okay so what I get is...

angle = 0.000000000592

so I use equation 3 and I get

cos(angle) = 0.999999999999999999824768

R - R(0.999999999999999999824768) =

1690000000000 - 1690000000000(0.999999999999999999824768) = 0.00000029614208

so... 2.96 x 10 ^ -7