00:01
We are given mass per unit length of the wire that is m by l that is given as 0 .5 gram per centimeter and current i is flowing through the wire that is i equal to 2 ampere and its direction is in south direction.
00:21
The current is flowing in the wire in the south direction.
00:25
Now we are going to find the magnetic field, minimum magnetic field, due to which the weight of the wire can be balanced.
00:40
Let the wire current is flowing in this direction in the wire i and we are going to find the value of minimum magnetic field that is required to lift this wire against.
00:56
The gravity so now let the magnetic field magnetic field is b is b so from f equal to i l cross b this is the formula that if a wire carrying a current i is placed in a magnetic field b then it experiences a force that is f equal to i l cross b now now this l, this is in south direction.
01:33
So we can take it as negative y -axis, that means in minus j -cap.
01:39
Now the gravity of this wire will be in the downward direction, will be in the downward direction.
01:48
Downward direction means we are talking about minus k -cap direction.
01:53
So i can write that weight of wire, weight of wire that is in minus k cap direction...