First slide

Magnetic force

Question

Proton with kinetic energy of 1 MeV 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 $1.6 \times 10^{- 27} k g)$

Moderate

A

$0.71 m T$
B

$0.071 m T$
C

$71 m T$
D

$7.1 m T$

Solution

$∵ K . E o f p r o t o n = \frac{1}{2} m V^{2} \Rightarrow 1 \times 10^{6} \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 1.6 \times 10^{- 27} \times V^{2}$
$\Rightarrow V = \sqrt{2} \times 10^{7} m / s$
$∴$ $M a g n e t i c F o r c e a c c e l e r a t e s t h e p r o t o n ∴ B q V = m a \Rightarrow B = \frac{m a}{q V} \Rightarrow B = \frac{1.6 \times 10^{- 27} \times 10^{12}}{1.6 \times 10^{- 19} \times \sqrt{2} \times 10^{7}}$
$= 0.71 \times 10^{- 3} T$
So, B = 0.71mT

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