NEET Practice Questions (MCQs) with Answers & Solutions

(a) As we know that, radius of a charged particle in a magnetic field B is given by r=$\frac{mv}{qB}$

 where,r= charge on the particle 

          v= speed of the particle

$\therefore$The time taken to complete the circle,

           $T=\frac{2\mathrm{\pi r}}{v}\Rightarrow \frac{T}{2\pi }=\frac{m}{qB}$      [from Eq. (i)]   

  $\therefore \omega =\frac{2\mathrm{\pi }}{T}=\frac{qB}{m}$

$\because q=e and \frac{e}{m}=1.76\times {10}^{11}C/kg$

             $$\begin{gathered} B=3.57\times {10}^{-2} T \\ \Rightarrow \frac{2\mathrm{\pi }}{T}=\frac{eB}{m} \\ f=\frac{1}{2\mathrm{\pi }}\frac{e}{m}B \left(\because \frac{1}{T}=f\right) \\ =\frac{1}{2\mathrm{\pi }}\times 1.76\times {10}^{-2} \\ = 1.0\times {10}^{9}Hz=1GHz \end{gathered}$$

Radius in magnetic fields of circular orbit,

R=mV/qB=√2mE/qB

and total energy of a moving particle in a circular orbit, E=q₂B₂R₂/2m

For a proton enter a region of magnetic field 
E₁=e₂ x B₂ x R₂/2xm_ρ  ...(i)

Where, is a mass of proton m_ρ
similarly for a α-particle moves in a uniform magnetic field 
E₂=(2e)² x B² X R²/2X(4m_ρ)  [∴m_α=4m_p]..(ii)

Dividing eq. (ii) by eq. (i) we get 
E₂/E₁=(2e)² x B² x R²/2X(4m_ρ) x 2 x m_ρ/^{e2 x B2 X R2}

E₂/E₁=1=> E₂=E₁=1MeV

In an ammeter with a shunt resistance, the shunt resistance allows most of the current to bypass the ammeter coil, reducing the potential difference across the coil. The ammeter reading is proportional to the potential difference across the coil. With the given resistances and voltage, the current through the circuit is 0.5 A, which corresponds to the ammeter reading.

The magnetic field at a point due to a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire. For two wires at right angles, the net magnetic field is the vector sum of the fields due to individual wires, which gives the magnitude as (μ0/2πd) * √(I1^2 + I2^2).

(b) Effective resistance,

R_eff=40x120/120+40=4800/160=30Ω

∴ Current I=7/(30+5)=7/35=0.2A

[∴ I=E/R+r]

(d) For parallel M is stable and for antiparallel is unstable.

The magnetic field (B) at the centre of circular current carrying coil of radius R and current I, B=$\frac{{\mu }_{0}I}{2R}$

Similarly, if current = 2I, then

             Magnetic field=$\frac{{\mu }_{0}2I}{2R}=2B$

So, resultant magnetic field

           =$\sqrt{B^2+{\left(2B\right)}^2}=\sqrt{5B^2}=\sqrt{5}B$

           =$\frac{{\mu }_{0}I\sqrt{5}}{2R}$

Frequency v=$\frac{eB}{2\mathrm{\pi m}}$

$KE=\frac{1}{2}m{v}^2 and radius R=\frac{mv}{eB}$

Here, velocity v=$\frac{\mathrm{\pi R}}{T/2}=\frac{2\mathrm{\pi R}}{T}=2\mathrm{\pi Rv}$

$\therefore Radius R=\frac{m\left(2\mathrm{\pi Rv}\right)}{eB}$

Magnetic field B=$\frac{2\mathrm{\pi mv}}{e}$

Kinetic energy

$K=\frac{1}{2}m{\left(2\mathrm{\pi Rv}\right)}^2=2m{\mathrm{\pi }}^2{\mathrm{v}}^2{\mathrm{R}}^2$

Magnetic field does not apply only force. The electric field will apply a force opposite to the velocity of the electron. Hence speed will decrease.