A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 6.
1 answer:
Answer:
The magnitude of the magnetic field is 10.19 T.
Explanation:
Given that,
Charge on the doubly charged molecule, q = 2e
Magnetic force,
Speed of molecule, v = 207 m/s
The angle to the direction of the field is 74.9 degrees.
We need to find the magnitude of magnetic field. The formula of magnetic force is given by :
B is magnetic field
So, the magnitude of the magnetic field is 10.19 T.
You might be interested in
Answer:
The magnitude of the force F is given by
F = ( +
+
) *(
*g/(
))
Explanation:
Given there are three blocks of masses ,
and
(ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( +
+
) *a ................(equation 1)
Also it is given that does not move with respect to
, which gives tension T is exerted on pulley by
only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T =
................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
*
=
* (
+
)
*
= (
*
)+ (
*
)
( -
) *
=
*
=
*
/(
-
)
Taking square root on both sides, we get acceleration a
a = *g/(
)
Hence substituting the value of a in equation 1, we get
F = ( +
+
) *(
*g/(
))
