00:01
In this problem an electron has velocity in positive x direction and acceleration in positive z direction and electric field is in positive z direction and we have to find out the y component of the magnetic field in the region.
00:21
Now velocity can return as magnitude of velocity which is speed times unit vector along velocity.
00:33
So unit vector will be i cap along positive x -axis.
00:39
Electric field is along z axis, so it can be written as electric field magnitude times k -cap, which is the unit vector along positive z direction.
00:54
Acceleration is also along positive z direction, so it can be written as magnitude of acceleration times k -cap, which is the unit vector along positive z direction.
01:09
Now, magnetic field b is equal to b1 icap plus b2, j cap plus b3 k cap.
01:22
B1, b2 and b3 are the three components along the x, y and z directions, and the value of these components is unknown.
01:33
Now we have to find the y component of the magnetic field.
01:37
So, the net force on the electron is equal.
01:43
To sum of electric force and magnetic force.
01:51
So, electric force plus magnetic force, this is the resultant force and it will be equal to mass times acceleration from newton's second law of motion.
02:07
Electric force is charge times electric field and magnetic force is charge times cross product of magnetic field and velocity q v cross b now electric field is e k cap v is v i cap cross magnetic field is b1 i cap plus b2 j cap plus b3 k cap and this is equal to mass times acceleration now q is common in both these terms so it can be taken out and we have e k cap plus cross product of velocity and magnetic field so first we find the cross product of the x component of field with the velocity so it will be vb1 i cap cross i cap plus v b2 i crap cross j cap plus v b3 i cap i cap cross k cap and this is equal to mass times acceleration now q can be taken to the right hand side and on the left hand side we have e k cap plus v b1 i cross i is zero v b2 i cross j is k cap plus v b3 i cross k is minus j and this will be equal to mass times mass upon charge times acceleration so now on the left hand side we have e k cap plus v b2 k cap minus v b3 j cap equal to mass upon charge and acceleration is along z axis so it is a k -k now the k components together e plus v b2 minus m upon q a k cap minus v b3 j cap equal to 0...