Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic torque & energyHard2 minPYQ_2020

PhysicsHardsingle choice

Proton with kinetic energy of $1 M e V$ moves from south to north. It gets an acceleration of $10^{12} m / s^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{- 27} k g$ )

Options:

Answer:

Solution:

$∴ K . E . = 1.6 \times 10^{- 13} = \frac{1}{2} \times 1.6 \times 10^{- 27} v^{2}$
$v = \sqrt{2} \times 10^{7} m / s$
$∴ B q v = m a$
$B = \frac{1.6 \times 10^{- 27} \times 10^{12}}{1.6 \times 10^{- 19} \times \sqrt{2} \times 10^{7}}$
$= 0.71 \times 10^{- 3} T$
So,

$B = 0.71 m T$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: PYQ_2020

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