Question

A proton enters in a magnetic field of flux density 2.5T with a velocity of 1.5x107ms-1 at an angle of

30owith the field find the force on the proton​

Answer 1

Answer:

3×10^-¹²

Explanation:

we know that force acting on any charge due to magnetic field is given as:

Fm=qvBsin¢

q=1.6×10-¹⁹ C

v=1.5×10⁷ ms-¹

B=2.5 T

thus Fm=1.6×10-¹⁹×1.5×10⁷×2.5×sin30

Fm=3×10-¹² N ( sin30=1/2)

Answer 2

Given:

A proton enters in a magnetic field of flux density 2.5 T with a velocity of 1.5x10⁷ m/s at an angle of

30° with the field.

To find:

Force on proton ?

Calculation:

• The general expression of force on a moving charge in an magnetic field is given as :

$F = q \times v \times B \times \sin( \theta)$

• 'q' is charge , 'v' is velocity , 'B' is field intensity, $\theta$ is angle between velocity vector and B vector.

$\rm \implies F = (1.6 \times {10}^{ - 19} )\times (1.5 \times {10}^{7} )\times 2.5 \times \sin( {30}^{ \circ} )$

$\rm \implies F = (1.6 \times {10}^{ - 19} )\times (1.5 \times {10}^{7} )\times 2.5 \times \dfrac{1}{2}$

$\rm \implies F = 3\times {10}^{ (- 19 + 7)}$

$\rm \implies F = 3\times {10}^{ - 12} \: N$

So, force on proton is 3 × 10^(-12) N.