PHY 104 Quiz on Magnetic Field and Forces
1)A solenoid is connected to a battery as shown in the figure to the right and a bar magnet is placed nearby. In what direction will the bar magnet move is response to the magnetic field created by the solenoid? Explain your answer by describing, in words, the magnetic polarity of the solenoid. Using the Right-hand Rule, the magnetic field of the solenoid will be directed towards the left as shown. This implies that the right end of the solenoid acts like a south pole of a magnet while the left end of the solenoid acts like a north pole (Rule: magnetic field lines exit the north pole and travel into the south pole).Thus the bar magnet to the right will feel an attraction to the apparent opposite pole of the solenoid adjacent to it. The bar magnet will move to the left towards the solenoid.
2)Use the relation and the definition of a Tesla (1T = 1), to demonstrate, by unit analysis, that the units for the permeability of free space 0 are .
First derive an expression for 0:
Next do the unit analysis
Thus the units of 0 are N/A2
3)An electron shoots into a confined uniform magnetic field directed into the page as shown in the figure. Its velocity is 3.72 x105 m/s directed horizontally as shown.
A)Which of the illustrated paths (a, b, c, or d) will the electron follow?
B)If the radius of curvature of the electron is 5.2 cm what is the strength of the magnetic field?
First set the magnetic force equal to the centripetal force as necessitated by circular motion principles.
Derive an equation for B and solve.
4)Carbon atoms from a prehistoric sample of charcoal found in an ancient cave are ionized by intense heating and injected,with various velocities, into a velocity selector/mass spectrometer illustrated to the right, for the purpose of measuring the 13C abundance relative to the 12C abundance to date the charcoal. The electric field and magnetic field strengths are given in the diagram.
A)What is the mass of the 13C atom?
B)What velocity ions will emerge from the velocity selector and enter the turning field? Derive the velocity selector relation by equating the downward electric force on the charges F = qE with the upward magnetic force on the charges F = qvB.
C)What will be the radius of curvature of the trajectory of a singly ionized 13C ion?
First set the magnetic force equal to the centripetal force as necessitated by circular motion principles.
Derive an equation for r and solve.
D)Will the radius of curvature of a 12C atom be greater or less than the 13C atom? By what percentage or fraction?
Since the radius of curvature of the trajectory is directly proportional to the mass of the ion, the 12C ion will have a smaller radius of curvature by a factor of or 92% smaller.
5)A long wire carrying a 5.00 A current penetrates into a region of a 0.44 T magnetic field as shown below. Using the dimensions in the figure determine the magnitude and direction of the net force on the wire.
Identify the directions of the forces consistent with the right-hand rule and use symmetry to eliminate any equal and opposite forces. You can see that the two opposing forces F1and F2 will cancel each other leaving only F3 as the sole contributor to the net force. The magnitude of F3 is given by
Thus the net force on the wire is 0.55 N directed to the right.
6)A current in a long straight wire produces a magnetic field of 8.0 T at 4.0 cm from the wire’s center.
A)What is the magnetic field strength 2.0 cm from the wire? Since , we see that B is inversely proportional to R. So by halving R you will double B. Thus, B = 16 T when R = 2 cm.
B)How far from the wire’s center will the magnetic field be 1.0 T? Again, since , we see that B is inversely proportional to R. So to decrease B by a factor of 8 you must increase R by the same factor. Thus, R= 8∙4 cm = 32 cm for B to equal 1.0 T.
C)What is the current in the wire?Solve for I.
7)A single circular loop 10.0 cm in diameter carries a 2.00 A current.
A)What is the magnetic field at the center of this loop?
B)Now connect 1,000 of these single loops in series within a length of 500 cm to create a 500 cm long solenoid. What is the magnetic field strength in the center of the solenoid?
C)Why is the magnetic field strength in the solenoid not 1,000 times the magnetic field strength of a single loop? (Answer in a few sentences.)
The ratio of the magnetic fields is . The reason the solenoid’s B field is not simply 1,000 times the B field of the single loop is that the loops of the solenoid are distributed over a 5 m length. Thus a single loop of the solenoid does not fully contribute to the B field at the center of another loop in the solenoid. The more distance between the two loops the less the contribution. Thus the 1,000 loops of the solenoid do not all equally of fully contribute to the magnetic field at a particular point in the center of the solenoid.