A proton of charge 1.6 x 10\(^{19}\)C is projected into a uniform magnetic field of flux density 5.0 x 10\(^5\)T. If the proton moves parallel to the field with a constant speed of 1.6 x 10\(^6\)ms\(^{-1}\), calculate the magnitude of the force exerted on it by the field

Astrobrain2684

1 year ago

o determine the magnitude of the force exerted on the proton by the magnetic field, we use the formula for the magnetic force on a moving charge:

𝐹
=
𝑞
𝑣
𝐵
sin
⁡
𝜃
Where:

𝐹
is the magnetic force,

𝑞
=
1.6
×
10
−
19
 
C
is the charge of the proton,

𝑣
=
1.6
×
10
6
 
ms
−
1
is the velocity of the proton,

𝐵
=
5.0
×
10
−
5
 
T
is the magnetic flux density,

𝜃
is the angle between the velocity of the proton and the magnetic field.

Step 1: Analyze the angle
Since the proton is moving parallel to the magnetic field, the angle
𝜃
between the velocity and the field is
0
∘
. For
𝜃
=
0
∘
,
sin
⁡
0
∘
=
0
.

Step 2: Calculate the force
Substitute the values into the formula:

𝐹
=
(
1.6
×
10
−
19
)
(
1.6
×
10
6
)
(
5.0
×
10
−
5
)
sin
⁡
0
∘
Because
sin
⁡
0
∘
=
0
:

𝐹
=
0
 
N
Final Answer:
The magnitude of the force exerted on the proton is 0.0 N, which corresponds to option A.

okraDEN2023

1 year ago

The question was not correctly typed, the magnetic flux density "B" was not given in the question

anotherniftEE

1 year ago

iN SIMPLE SENSE
If its not perpendicular the V is zero and so the forece is zero, so anything parallel is a trick bcuz it is zero

Jaymyma001

1 year ago

There is no flux density!!!