This BITSAT test consists of BITSAT electromagnetic induction questions which are part of the BITSAT Physics section. Get detailed information related to BITSAT sample papers, BITSAT hall ticket & BITSAT cut off.
End Test Now
A car moves on a plane road. The emf induced in the axle connecting the wheels is maximum, when it moves (ignore earth’s rotation)
westward at the equator
eastward at the equator
eastward at the latitude of 45°
at the poles
A coil of area 500 and 1000 turns is put perpendicular to a magnetic field of 0.4 gauss. If it is rotated by in 0.1 second, the emf induced will be
20 mV
40 mV
0 mV
60 mV
A long and straight wire lies along the axis of a straight solenoid, as shown. If the wire carries a current , the emf e induced in the solenoid will be
zero
When the circuit of outer coil P, shown in the figure, is closed the current induced in the inner coil Q will be
Clockwise and instantaneous
Anticlockwise and continuous
Anticlockwise and instantaneous
Clockwise and continuous
The rotational emf induced in a wire moving in a magnetic field does not depend on it’s
Orientation
Velocity
Diameter
Length
A rectangular conducting bar of breadth w and infinite length is placed along the x-axis, as shown, in a magnetic field acting in the y-direction. A voltmeter is placed with the two wires in contact at two points across the breadth. On moving the wires, carrying the voltmeter with the speed v, the reading of voltmeter will be
vwB
A metal rod of length moves with a constant velocity v in a direction perpendicular to itself. A constant and uniform magnetic field exists in space in a direction perpendicular to the rod as well as velocity. Select the correct statement
The electric field will be induced in the rod
The electric potential will be highest at the centre of rod and decreases towards the ends
The entire rod will be at the same potential
The electric potential will be lowest at the centre of rod and increases towards the ends
A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic field . At the position MNQ of ring, the speed of ring is v. Then, the potential difference developed across the ring will be
Zero
2RvB and point Q will be at higher potential
PQ is a straight wire, carrying an electric current due north. A conducting loop abcda is moving with the velocity v towards east. Then, the current induced in the loop will be
Clockwise
First anticlockwise and then, clockwise
Anticlockwise
First clockwise and then. Anticlockwise
When you are sure that you have answered as many questions as possible, click the ‘Done’ button below and view your results.