Q2.The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. what will be its value at the centre of the loop?
Answer : Option DExplaination / Solution: The magnetic field at the centre of a coil of radius R and number of turns N , carrying a current I is B0=μ0NI2R. At a point distance x from the coil, the field is Bx=μ0NIR22(R2+x2)32
Q3.A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the center of the coil such that r>>R, varies as
Answer : Option BExplaination / Solution: At a point distance r from the coil, the magnetic field is Br=μ0NIR22(R2+r2)12. If r>>RR is neglected in the denominatorBr=μ0NIR2r3;B∝1r3
Q4.A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be
Answer : Option CExplaination / Solution: If the length of the wire is L and the radii of the coils in two cases be R1 and R2. Then,
Q6.A magnetic needle lying parallel to a magnetic field requires W units of work to turn it, through 600. The torque needed to maintain the needle in this position will be
Answer : Option BExplaination / Solution: the work done to turn a needle through an angle θ is W=mBcosθThe torque needed to maintain τ=mBsinθ.τW=tanθ.τ=Wtanθ=Wtan60=3–√W
Q7.If number of turns, area and current through a coil is given by n, A and i respectively then its magnetic moment will be
Answer : Option DExplaination / Solution:
The magnetic moment associated with a coil carrying current is given by the product of its area and the current through it. M=n I A
Q8.The work done in rotating a magnet of magnetic moment 2A−m2 in a magnetic field of 5×10−3T form the direction along the magnetic field to opposite direction to the magnetic field, is
Answer : Option AExplaination / Solution: The potential energy of a magnetic dipole of moment m placed in a magnetic field is U=−mBcosθWhen the magnet is aligned in the direction of the field, and the initial potential energy Ui=−mBWhen the magnet is aligned opposite to the direction of the fieldθ=180its potential energy isUf=mBWork done in rotating the magnet is equal to the change in its potential energy.
Q9.A magnet of magnetic moment 2JT−1 is aligned in the direction of magnetic field of 0.1.T. What is the net work done to bring the magnet normal to the magnetic field?
Answer : Option BExplaination / Solution: The potential energy of a magnetic dipole of moment m placed in a magnetic field is U=−mBcosθ.When the magnet is aligned in the direction of the field, and the initial potential energy Ui=−mBWhen the magnet is placed perpendicular to the direction of the field,θ=90its potential energy is Uf=0.Work done in rotating the magnet is equal to the change in its potential energy. W=Uf−Ui=0−(−mB)=mB=2×0.1=0.2J
Answer : Option AExplaination / Solution:
A solenoid carrying current produces a magnetic field very similar to that of bar magnet. The magnetic field lines emerge from the ends of a solenoid and the number of field lines near its perpendicular bisector is almost equal to zero. A circular coil produces field along its axis. A straight conductor produces a magnetic field that can be represented by concentric circles. A toroid is a solenoid that has collapsed on itself. The field in a toroid is confined to the ring like region bounded by the toroid.
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0