To solve the problem of finding the ratio of the radius of the deuteron path (\( r_d \)) to the radius of the proton path (\( r_p \)) in a uniform magnetic field, we follow a step-by-step approach:
The radius \( r \) of a charged particle's path moving perpendicular to a magnetic field is given by the expression: \(r = \frac{mv}{qB}\) where:
The kinetic energy \( KE \) for both particles is:
\(KE = \frac{1}{2}mv^2\)
Given that their kinetic energies are equal, set the kinetic energy equations equal and solve for velocity:
\(\frac{1}{2}m_pv_p^2 = \frac{1}{2}m_dv_d^2\)
where:
From here, find the relationship between their velocities:
\(v_p^2 = \frac{m_d}{m_p} v_d^2\)
Thus, \(v_p = \sqrt{\frac{m_d}{m_p}} v_d\)
For the proton, \(r_p = \frac{m_pv_p}{q_pB}\)
For the deuteron, \(r_d = \frac{m_dv_d}{q_dB}\)
The charge \(q = +e\) for both particles, so it cancels out when comparing
The ratio of the radii is given by:
\(\frac{r_d}{r_p} = \frac{\left(\frac{m_dv_d}{qB}\right)}{\left(\frac{m_pv_p}{qB}\right)} = \frac{m_d v_d}{m_p v_p}\)
Substitute \(v_p = \sqrt{\frac{m_d}{m_p}} v_d\):
\(\frac{r_d}{r_p} = \frac{m_d v_d}{m_p \left(\sqrt{\frac{m_d}{m_p}} v_d\right)} = \sqrt{\frac{m_d}{m_p}}\)
Given that \(m_d = 2m_p\) (for a deuteron which is made up of a proton and a neutron), we have:
\(\frac{r_d}{r_p} = \sqrt{2}\)
The ratio of the radii of the paths of the deuteron and proton is \(\sqrt{2} : 1\). Thus, the correct answer is \( \sqrt{2} : 1 \).
The radius of the path of a charged particle moving in a magnetic field is given by:
\( r = \frac{mv}{qB} \),
where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.
Since both particles have the same kinetic energy, \( \frac{1}{2}mv^2 = K \), and the velocity \( v \) can be expressed as:
\( v = \sqrt{\frac{2K}{m}} \).
Thus, the radius of the proton path \( r_p \) is:
\( r_p = \frac{m_p \sqrt{2K/m_p}}{eB} \),
and the radius of the deuteron path \( r_d \) is:
\( r_d = \frac{m_d \sqrt{2K/m_d}}{eB} \).
Since \( m_d = 2m_p \), the ratio of the radii is:
\( \frac{r_d}{r_p} = \frac{\sqrt{2m_p}}{\sqrt{m_p}} = \sqrt{2} \).
Thus, the ratio is \( \sqrt{2} : 1 \), and the correct answer is Option (3).
A current-carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
