Question

(5)
An electric charge of 35uC is moving with speed 2x10ms along a path shown in figure.
Then magnetic field produced at point P is
ut.
.P
(A) Zero
(B) 242.5
r
50 mm
(C) 2425
(D) 2524
q=35uc
60°
y
I​

Answer 1

Answer:

) Let

B

=B

0

(−

K

^

)

V

=V(−

x

^

)

Electromagnetic force F

m

=q

V

×

B

=qVB(−

j

^

)

Now we can conclude that point charge move on circular motion of radius r=

qB

mV

and motion of charge in X−Y plane.

(ii) If point charge has velocity component parallel to

B

then, In this case path followed by charge particle is helical path.

(iii) If electric field is also applied such that the particle continuous moving along straight line the electromagnetic force should be balanced by electrostatics force F

e

F

m

=qVB(−

j

^

)

If net force on q is zero.

F

e

=qV

j

^

q

E

=qVB(

j

^

)

E

=VB(

j

^

)