Question

A proton enters a uniform magnetic field of 5T with velocity 4×10^7 ms^-1 at right angles to the field.The magnetic force on the proton is(Charge on a proton = 1.6×10^-19C)

Answer 1

force on a proton is given by the formula

F = q*v*B

F = 1.6 x 10⁻¹⁹ x 4 x 10⁷ x 5

  = 32 x 10⁻¹³

  = 3.2 x 10⁻¹² newton

the magnetic force on that proton = 3.2 * 10¹² N

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Answer 2

Answer:

Magnetic force, F = 6.4 × 10⁻¹² N

Explanation:

It is given that,

Charge of proton, $q=1.6\times 10^{-19}\ C$

Magnetic field, B = 5 T

Velocity of the proton, $v=4\times 10^7\ m/s$

Magnetic force is given by :

$F=qvBsin\ \theta$

Putting all the values in the above equation :

$F=1.6\times 10^{-19}\ C\times 4\times 10^7\ m/s\times 5\ T$

$F=3.2\times 10^{-11}\ N$

Hence, this is the required solution.