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Chapter 15: Problem 28

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

Short Answer

Expert verified
The ratio of masses of the two particles X and Y is given by: (A) \( \left(\frac{R_{1}}{R_{2}}\right)^{1 / 2} \)

Step by step solution

01

Recall the Centripetal force formula

In a magnetic field, a charged particle experiences a force that makes it move in a circle. The centripetal force formula is given by: \( F = \frac{mv^2}{r} \) where \(F\) is the centripetal force, \(m\) is the mass of the particle, \(v\) is its velocity, and \(r\) is the radius of the circular path.
02

Recall the Lorentz force formula

The force acting on a charged particle in a magnetic field is known as the Lorentz force. The equation is given by: \( F = qvB \) where \(F\) is the Lorentz force, \(q\) is the charge of the particle, \(v\) is its velocity, and \(B\) is the magnetic field strength.
03

Equate the centripetal force and Lorentz force formulas

Since the centripetal force is equal to the Lorentz force in this case, we set the two equations we derived earlier equal to each other. \( \frac{mv^2}{r} = qvB \)
04

Solve for the radius

We now want to isolate the radius, \(r\), to create an equation that relates the radii to masses of the two particles. We do this by dividing both sides by \(v\), which gives us: \( \frac{mv}{r} = qB \) Now, solving for the radius, \(r\), we get: \( r = \frac{mv}{qB} \)
05

Apply the equation to both particles

Now, we can apply this equation to both particles X and Y: For particle X: \( R_1 = \frac{m_1v_1}{qB} \) For particle Y: \( R_2 = \frac{m_2v_2}{qB} \) Since both particles are accelerated through the same potential difference, their kinetic energies are equal, which means: \( \frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2 \)
06

Find the ratio of the masses

Now we want to find the ratio of the masses \(\frac{m_1}{m_2}\). First, divide the equation for particle X by the equation for particle Y: \( \frac{R_1}{R_2} = \frac{m_1v_1}{m_2v_2} \) Now, recall the kinetic energy equality mentioned earlier: \( \frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2 \) Divide both sides of the equation by \(\frac{1}{2}v_1^2\) and then by \(\frac{1}{2}v_2^2\) to get: \( \frac{m_1}{m_2} = \frac{v_2^2}{v_1^2} \) Now, substitute this back into the ratio equation derived earlier: \( \frac{R_1}{R_2} = \frac{v_2^2}{v_1^2} \) Square rooting both sides, we find the ratio of the masses: \( \sqrt{\frac{R_1}{R_2}} = \frac{m_1}{m_2} \) After analyzing the answer choices, we can see that the correct option is: (A) \( \left(\frac{R_{1}}{R_{2}}\right)^{1 / 2} \)

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Most popular questions from this chapter

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\). Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3}\). The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

Two long parallel wires carrying equal currents in opposite directions are placed at \(x=\pm a\) parallel to \(v\)-axis with \(z=0\). Then (A) Magnetic field \(B_{1}\) at 1\. \(\frac{\mu_{0} i}{3 \pi a}\) origin \(O\) (B) Magnetic field \(B_{2}\) at \(P\) 2\. \(\frac{\mu_{0} i}{4 \pi a}\) \((2 a, 0,0)\) \((a, 0,0)\) (C) Magnetic field at \(M\) 3\. \(\frac{\mu_{0} i}{\pi a}\) (D) If wire carries current in 4\. Zero the same direction, then magnetic field at origin 5\. None

Three infinitely long thin conductors are joined at the origin of coordinates and lies along the \(x, y\), and \(z\) axes. A current \(i\) flowing along the conductor lying along the negative \(x\)-axis divides equally into the other two at the origin. The magnitude of magnetic filed at the point \((0,-a, 0)\) is (A) \(\frac{\mu_{0} i}{4 \pi a}\) (B) \(\frac{3 \mu_{0} i}{4 \sqrt{2} \pi a}\) (C) \(\frac{\sqrt{5} \mu_{0} i}{8 \pi a}\) (D) \(\frac{\sqrt{3} \mu_{0} i}{2 \pi a}\)

Two short bar magnets of length \(1 \mathrm{~cm}\) each have magnetic moments \(1.20 \mathrm{Am}^{2}\) and \(1.00 \mathrm{Am}^{2}\), respectively. They are placed on a horizontal table parallel to each other with their \(\mathrm{N}\) poles pointing towards the south. They have a common magnetic equator and are separated by a distance of \(20.0 \mathrm{~cm}\). The value of resultant horizontal magnetic induction at the midpoint \(O\) of the line joining their centres is close to [2013] (Horizontal component of earth's magnetic induction is \(\left.3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\right)\) (A) \(2.56 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(3.50 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(5.80 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\)

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.
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