Answer : Option CExplaination / Solution:
For measurement of current of any magnitude, the galvanometer is always connected in series. This is to ensure that the entire current flows through the galvanometer. If the galvanometer were connected in parallel, the current branches out and only a part of the current will flow through it.
Answer : Option CExplaination / Solution:
A shunt resistance of magnitude less than that of the galvanometer is added in parallel to the galvanometer to convert it into an ammeter. The equivalent resistance of a parallel combination is always less than the least value of the resistances in parallel. Ammeters are connected in series to the circuit to allow the entire current of the circuit to pass through them. Its resistance has to be very less- almost negligible, so that it does add on to the resistance in the circuit, thereby changing the current.
Q3.A Current carrying wire produces in its neighbourhood
Answer : Option DExplaination / Solution:
Static charges produce only an electric field. Moving charges produce a magnetic field in addition to their electric fields.
Q6.The magnetic field due to very long wire carrying current varies according to
Answer : Option CExplaination / Solution: The magnetic field at a point, which is at a distance r, from a very long wire carrying current is B=μ0I2πr B∝1r
Q7.What should be the current in a circular coil of radius 5 cm to annul BH=5×10−5 T at the center?
Answer : Option CExplaination / Solution: The magnetic field produced by the coil at the centre should be equal and opposite to the BHThe field at the centre of a circular coil B=μ0I2r.The current needed to produce a field equal to BH
Q10.A particle having charge 100 times that of an electron is revolving in a circular path of radius 0.8 m with one rotation per second. Magnetic field produced at the centre of the circular path is
Answer : Option CExplaination / Solution: A charge moving in a circular path is equivalent to a current I=qTSince the particle has charge 100 times e and it makes 1 revolution per second,q=100e and T=1s.
The magnetic field at the centreB=μ0I2r=μ0(1.6×10−17)2×0.8=μ0×10−17
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0