Question

A proton enters a uniform magnetic field of 10T with velocity 7 1 2 10 ms−

× at right

angles to the field. The magnetic force acting on the proton is ?​

Answer 1

Explanation:

Angle between velocity of proton and magnetic field θ=30

o

Velocity of proton v=2×20

7

m/s

Magnetic field B=1.5Wb/m

2

Magnetic force acting on the proton F=qvBsinθ

∴ F=(1.6×10

−19

)(2×10

7

)(1.5)×0.5

⟹ F=2.4×10

−12

N

Answer 2

Answer:

The magnetic force acting on the proton is 1.12×10⁻⁵N.

Explanation:

The magnetic force is calculated as,

$\mathrm{F=evB}$                (1)

Where,

F=force due to magnetic field

e=charge on an electron

v=velocity with which the proton is moving

B=the magnetic field in which the charge is moving

From the question we have,

The magnetic field=10 T

The velocity with which the proton is moving=7×10¹² m/s

The charge on a proton=1.6×10⁻¹⁹Coloumb

By inserting the numbers in equation (1), we receive;

$\mathrm{F=1.6\times 10^{-19}\times 7\times 10^{12}\times 10}$

$\mathrm{F=1.12\times 10^{-5}\ N}$

Hence, the magnetic force acting on the proton is 1.12×10⁻⁵N.

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