To solve this we need to use 2 right hand rules, one to determine the direction of the magnetic field at the position of one of the wires, due to the other wire, and once we have this we need to determine the direction of the force.
CloseIf the particle would have been shot in at a right angle to the field lines, then C would be correct. However, here the particle also has an initial velocity to the right, so it will make circles, all while moving to the right, which has the shape of a spiral/helix.
CloseIn this case, we have two forces acting on the particle which cancel each other out since we have constant speed. One of these forces is the magnetic force, the other one is the electric force. So we have \(qvB = Eq\), rearranging for \(v\) gives \(v = \frac{E}{B}\).
CloseWe need to use the right-hand rule. The field points from the top down since magnetic field lines point from north to south outside the magnet, and the speed of the electron is to the right. We use our hand to find the force, but we need to make sure we take the opposite to what our hand says as that would be the force for a proton, but we have an electron. Hence, the correct answer is B.
CloseThe force acting on the particle is:
\[ F = qE \]
Substituting \( q = 1.5 \times 10^{-6} \: \mathrm{C} \) and \( E = 5000 \: \mathrm{N/C} \):
\[ F = (1.5 \times 10^{-6})(5000) = 7.5 \times 10^{-3} \: \mathrm{N} \]
The acceleration is given by:
\[ a = \frac{F}{m} = \frac{7.5 \times 10^{-3}}{2.0 \times 10^{-3}} = 3.75 \: \mathrm{m/s^2} \]
CloseThe magnetic force is given by:
\[ F = qvB \sin\theta \]
Here, \( \theta = 90^\circ \), so \( \sin\theta = 1 \). Substituting:
\[ F = (2.0 \times 10^{-6})(300)(0.5) = 3.0 \times 10^{-4} \: \mathrm{N} \]
CloseFor a charged particle to pass through undeflected, the electric and magnetic forces must balance:
\[ qE = qvB \implies v = \frac{E}{B} \]
Substituting \( E = 4000 \: \mathrm{N/C} \) and \( B = 0.2 \: \mathrm{T} \):
\[ v = \frac{4000}{0.2} = 20000 \: \mathrm{m/s} \]
CloseThe magnetic force on a current-carrying wire is given by:
\[ F = ILB \sin\theta \]
Here, \( \theta = 90^\circ \), so \( \sin\theta = 1 \). Substituting:
\[ F = (10)(0.5)(0.3) = 1.5 \: \mathrm{N} \]
CloseThe radius of the circular motion is given by:
\[ r = \frac{mv}{qB} \]
Substituting \( m = 1.67 \times 10^{-27} \), \( v = 1.0 \times 10^6 \), \( q = 1.6 \times 10^{-19} \), and \( B = 0.2 \):
\[ r = \frac{(1.67 \times 10^{-27})(1.0 \times 10^6)}{(1.6 \times 10^{-19})(0.2)} = 0.05 \: \mathrm{m} \]
CloseThe work done by the electric field is:
\[ W = qEd \]
Substituting \( q = 2.0 \times 10^{-6} \), \( E = 1000 \), and \( d = 0.5 \):
\[ W = (2.0 \times 10^{-6})(1000)(0.5) = 0.001 \: \mathrm{J} \]
CloseThe direction of the magnetic force is determined by the right-hand rule:
Thus, the force is directed up.
CloseThe magnetic force on a current-carrying wire is given by:
\[ F = ILB \sin\theta \]
Here, \( I = 5 \: \mathrm{A} \), \( L = 0.8 \: \mathrm{m} \), \( B = 0.4 \: \mathrm{T} \), and \( \theta = 30^\circ \). Plugging in the values:
\[ F = (5)(0.8)(0.4)(\sin 30^\circ) = (5)(0.8)(0.4)(0.5) = 0.8 \: \mathrm{N} \]
Close