Explanation
The number of turns in two concentric coils are n1andn2 and the ratio of their radii is 2:When equal currents is passed through them in opposite direction, the magnetic field produced at the center becomes zero. The ratio n1:n2 will be:
Given: Length l and magnetic moment M of steel wire
To find: New magnetic moment
Solution:
We know, pole strength is given by,
m=Ml
A steel wire is bent into semi-circular arc, so if r is radius of semicircle, then
πr=l,
Therefore, r= 1π
Now new magnetic moment M′=m×2r
=Ml×2lπ
=2Mπ =2Mπ−1
Hence B is the correct option
Correct Answer: A
Hint: Calculate net magnetic field direction at R and then use →F=i(→dl×→B)
Step-1: Force on R due to the current in P & Q
∴ Force per unit length on the wire R due to Q wire is
F1=μoI3I22πr=24μo2πr direction towards left according to Fleming's left hand rule
Similarly, force per unit length on the wire R due to P wire is
F2 = μoI1I32π×2r = 12μo4πr direction towards left according to Fleming's left hand rule
Step-2: Resultant force
∴ Resultant force F per unit length = F1+F2
=24μo2πr+12μo4πr
=60μo4πr=15μoπr
∵ F1 and F2 both acting towards left therefore resultant force also acting towards left on the wire R
Hint: The resultant magnetic field is the vector sum of Bx and By .
Correct Option: C
Explanation for Correct Option:
Step 1: Find the magnetic field due to the half ring wire,
Magnetic field due to the half ring wire, B1=μoi4R ,
where i is the current through the wire and R is the radius of the wire.
Step 2: Magnetic Field induction at point O due to the two straight wires,
Magnetic field at O, B2=2[μoi4πR(cos0−cos90)]
Step 3: Find the resultant magnetic field at point O.
Resultant magnetic field,
B=√B12+B22
B=μoi4πR(π2+22)12
So, the magnetic induction at point P is B=μoi4πR(π2+22)12
Two circular concentric loops of radii r1=20cm and r2=30cm are placed in the X-Y plane as shown in the figure. A current I=7 A is flowing through them. The magnetic moment of this loop system is
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