Solveeit Logo
Login

Question

Physics Question on Magnetic Field

A proton of mass 1.67×10−27 kg 1.67 \times 10^{-27}\, kg enters a uniform magnetic field 1 T1\, T of at point A shown in figure with a speed of 107 ms−110^7\, ms^{-1} The magnetic field is directed normal to the plane of paper downwards. The proton emerges out of the magnetic field at point CC, then the distance ACAC and the value of angle θ\theta will respectively be

A

0.7 m, 45∘0.7\, m ,\, 45^{\circ}

B

0.7 m, 90∘0.7\, m ,\, 90^{\circ}

C

0.14 m, 90∘0.14\, m ,\, 90^{\circ}

D

0.14 m, 45∘0.14\, m ,\, 45^{\circ}

Answer

0.14 m, 45∘0.14\, m ,\, 45^{\circ}

Explanation

Solution

From the symmetry of figure, the angle θ=45∘\theta=45^{\circ}.
The path of moving proton in a normal magnetic field is circular. If rr is the radius of the circular path, then from the figure
AC−2rcos⁡45∘−2r×12−2rA C-2 r \cos 45^{\circ}-2 r \times \frac{1}{\sqrt{2}}-\sqrt{2} r ...(i)
As Bqv=mv2rB q v=\frac{m v^{2}}{r} or r=mvBq r=\frac{m v}{B q}
AC=2mvBq=2×1.67×10−27×1071×1.6×10−19A C =\frac{\sqrt{2} m v}{B q}=\frac{\sqrt{2} \times 1.67 \times 10^{-27} \times 10^{7}}{1 \times 1.6 \times 10^{-19}}
=0.14 m=0.14\, m