The figure shows a region of length ‘ l ‘ with a uniform magnetic field of 0.3 T
Question:

The figure shows a region of length ‘ $l$ ‘ with a uniform magnetic field of $0.3 \mathrm{~T}$ in it and a proton entering the region with velocity $4 \times 10^{5} \mathrm{~ms}^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘ $l$ ‘ is close to (mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.6 \times 10^{-19} \mathrm{C}$ )

  1. (1) $0.11 \mathrm{~m}$

  2. (2) $0.88 \mathrm{~m}$

  3. (3) $0.44 \mathrm{~m}$

  4. (4) $0.22 \mathrm{~m}$


Correct Option: , 3

Solution:

(3) Time period of one revolution of proton, $T=\frac{2 \pi m}{q B}$

Here, $m=$ mass of proton

$q=$ charge of proton

$B=$ magnetic field.

Linear distance travelled in one revolution, $p=T(v \cos \theta)$ (Here, $v=$ velocity of proton)

$\therefore$ Length of region, $l=10 \times(v \cos \theta) T$

$\Rightarrow l=10 \times v \cos 60^{\circ} \times \frac{2 \pi m}{q B}$

$\Rightarrow l=\frac{20 \pi m v}{q B}=\frac{20 \times 3.14 \times 1.67 \times 10^{-27} \times 4 \times 10^{5}}{1.6 \times 10^{-19} \times 0.3}$

$\Rightarrow l=0.44 \mathrm{~m}$

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