In: Physics

# For the conditions given in the previous question, the charge's actual trajectory is:

In the figure below, a magnetic field of 0.6 Tesla points into the page. A particle of charge $$-12 n C$$ and mass $$0.6 ? g$$ is shot into the field as shown in the figure with a velocity of $$0.1 \mathrm{~km} / \mathrm{s} .$$ What is the radius of the orbit (in meters)? (The radius is a positive quantity.) Be careful with the powers of 10. Hint: k=kilo, m=milli, ?=micro, $$n=$$ nano and the SI units for B-field, charge, mass, length and time are: tesla (T), coulomb (C), kilogram (kg), meter (m) and second (s), respectively.

For the conditions given in the previous question, the charge's actual trajectory is:
A. a circle bending to the left
B. a parabola bending to the left
C. a circle bending to the right
D. a parabola bending to the right

## Solutions

##### Expert Solution

(1) When the charge particle enters perpendicular to the magnetic field it follows a circular path. The centripetal force acting on the particle is equal to the force acting on the particle due to magnetic field. The centripetal force acting on the charge is, $$F_{\mathrm{c}}=\frac{m v^{2}}{r}$$ The force acting on the particle due to magnetic field is, $$F_{\mathrm{b}}=q v B \sin 90^{\circ}$$ $$=q v B$$ The radius of the orbit is calculated as follows: \begin{aligned} F_{\mathrm{b}} &=F_{\mathrm{c}} \\ q v B &=\frac{m v^{2}}{r} \\ r &=\frac{m v}{q B} \\ &=\frac{\left(0.6 \times 10^{-6} \mathrm{~g}\left(\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~g}}\right)\right)\left(0.1 \times 10^{3} \mathrm{~m} / \mathrm{s}\right)}{\left(12 \times 10^{-9} \mathrm{C}\right)(0.6 \mathrm{~T})} \\ &=8.333 \mathrm{~m} \end{aligned}

(2) The force acting on the negative charge is pointing towards the positive $$x$$ axis. Hence, the trajectory is a circle bending to the right. Hence, the correct option is (c).

(1) =8.333

(2)  the correct option is (c).

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