© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Problem
A proton with kinetic energy of 1 MeV is moving from south to north. A magnetic field is applied from west to east, causing the proton to experience a given acceleration of $10^{12}\,\text{m/s}^2.$ We need to find the magnitude of this magnetic field.
Step 2: List the Known Quantities
Kinetic energy of proton, $K = 1\,\text{MeV} = 1 \times 10^6 \,\text{eV}$
$1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$
Acceleration of proton, $a = 10^{12}\,\text{m/s}^2$
Rest mass of proton, $m = 1.6 \times 10^{-27}\,\text{kg}$
Charge of proton, $q = 1.6 \times 10^{-19}\,\text{C}$
Step 3: Convert Kinetic Energy to Joules
$K = 1\,\text{MeV} = 1 \times 10^6 \,\text{eV} \times 1.602 \times 10^{-19}\,\text{J/eV} = 1.602 \times 10^{-13}\,\text{J}.$
Step 4: Find the Velocity of the Proton
The kinetic energy of the proton is given by
$K = \frac{1}{2}mv^2.$
Solving for $v$:
$v = \sqrt{\frac{2K}{m}}
= \sqrt{\frac{2 \times 1.602 \times 10^{-13}}{1.6 \times 10^{-27}}}
= \sqrt{2.0025 \times 10^{14}}
\approx 1.416 \times 10^{7}\,\text{m/s}.$
Step 5: Relate Force to Magnetic Field
The net force required to produce the proton's given acceleration is
$F = m \, a = (1.6 \times 10^{-27}\,\text{kg}) \times 10^{12}\,\text{m/s}^2
= 1.6 \times 10^{-15}\,\text{N}.$
The magnetic force acting on a moving charge is
$F = q v B.$
Since the velocity is perpendicular to the magnetic field, we can write
$m a = q v B
\quad\Longrightarrow\quad
B = \frac{m a}{q v}.
Step 6: Calculate the Magnetic Field
$B
= \frac{1.6 \times 10^{-15}}{\left(1.6 \times 10^{-19}\right)\,(1.416 \times 10^7)}
= \frac{1.6 \times 10^{-15}}{2.2672 \times 10^{-12}}
\approx 0.705 \times 10^{-3}\,\text{T}
= 0.705\,\text{mT}
\approx 0.71\,\text{mT}.
Step 7: Final Answer
The required magnetic field is approximately $0.71\,\text{mT}.$
