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_·-c- _. .... _
and the horizontal drawn in the magnetic meridian. (lmark)
Chapter 15: Mechanical Effect of Electric Current
SI. QuestionNo.
1. Force on a charged particle moving in a magnetic field is given by
a) F = Bq v cos 0
b) F = B q v sin {)
c) F= Bq cosOv
d) F = Bq cosOv
Ans: b)
2. The force on a charged particle moving in a magnetic field IS Kmaximum at
a) 9=0
b) 9=450
c) 9 = 90°
d) 0=1800
Ans : c)
Obj/ Spec.!
Diff. Level
U
See
relationship,
generalize
Average
Recall
Easy
232
5/13/2018 15.Mechanical Effect of Electric Current (Physics - Chapter Wise Question Bank...
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I I P UC
3. A charged particle at rest is placed in a magnetic field experiences K
zero force. Why? Recall
Easy
F-B q v sin eAtrestv=O,:. F=O (1 mark)
4. State Fleming's left hand rule. K
State
Easy
Statement ~ (1 Mark)
Show directions of Force field and velocity (1 mark)
5. Mention an expression for the torque acting on a current loop K
placed in a uniform magnetic field. Recall
Easy't = M B cos a (lmark)
6. Mention an expression for the torque on a current loop placed in a Kuniform magnetic field with the normal to the plane of the coil Recall
making an angle awith the direction of the field. Easy
1: = MB sin a (1mark)
7. Write the relation for the force on a current-carrying conductor kept Kin a magnetic field. Recall
Easy
F = B I I (1mark)
Or F =B 11sin 8
8. Briefly mention how a galvanometer can be converted to an K
ammeter. Recall,
Recognise
Average
By connecting a small resistance in parallel with galvanometer.
(1 mark)
S = Ig G
I~ Ig
(1 mark)
9. Briefly mention how a galvanometer can be converted to a volt K
meter. Recall and
express
233
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I I PUC
Average
By connecting a high resistance in series with galvanometer.
(1mark)
VR = - - G (I mark)
If f
10 Arrive at an expression for the force between two parallel U
conductors carrying currents. Explain and
establish
Average
Field on second conductor due to current IIof first conductor
B = Po 1\ (1mark)2 1 C a
Force F =BhI (1mark)
F =Po II X hi (1 mark)2 1 C O
F - Po I, 12 1 k)1- (mar
2 1 C a
II Describe with theory the working of a moving coil galvanometer. K
Describe,
locate,
express
AverageDiagram (1 mark)
Description of working (1 mark)
C = N B II x b =NB I A (I mark)
Restoring couple = deflecting couple
C, = DD (l mark)
Showing I oc e (1mark)
12 Describe an experiment to determine the current sensitivity of a U
pointer galvanometer. Recall,
express,
tabulate
Average
Diagrams (1 mark)
Formula
C'.. d (P + Q) R x 10--( ; d i .' 11k
urrent sensitivity = rvisions )lA ( mar)EQ
Procedure in brief (2 marks)
Tabular columns (lmark)
234
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[[PUC
13 Calculate the current sensitivity of the pointer galvanometer using U
the following data. Compute
EMF of the cell = 2V. Average
Trial Resistance Current in one Current in the
No. unplugged (ohms) direction (A) opposite direction
(A)
L 5000 5000 23 4000 25 4100
2. 6000 4000 19 4100 19 4000
3. 7000 3000 15 4050 14 4050
Trial No.1.
Formula
Mean deflection = 24 div. }Mean R = 4050 nCurrent sensitivity = 0.097 divl J .l
(1 mark)
(1 mark)
Trial No.2:
Mean deflection =19 div.
Mean R=4050 1 1 :
Current Sensitivity = 0.096 div / J.lA} (Imark)
Trial No.3.
Mean deflection = 14.5 div
Mean R = 4050 0. } (1 mark)Current sensitivity = 0.098 div I J . l A
Overall - (1 mark)
14 Describe an experiment to convert a galvanometer into a voltmeter. U
Recall,
explain
AverageDiagram (1 mark)
VFormula R=- - G (1mark)
1 8
Procedure in brief (2 marks)Tabular column (1 mark)
IS Calculate the resistance to be connected in series with the given U
galvanometer to convert it into a voltmeter using the following data. Compute
Average
235
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[[PUC
Resistance of the galvanometer = 200 Q
Current sensitivity of the galvanometer = 5 div / 1 - 1 A
No. of div, On one side of the end of the galvanometer = 30 div.Range of the voltmeter = 0 to 10 V.
VR=- -G (1mark)
1 9
10= J - 200 = 1800Q (1mark)
5 x 10-
16 Describe an experiment to determine the value of Bli at a place U
using a tangent galvanometer. Recall,
Express,
ExplainAverage
Diagram (1 mark)
Formula BH = J 10
n K }2r
JK= --
tan B
(lmark)
Procedure (2 marks)
Tabular column (1 mark)
17. Compute the value of BHat a place using following data.
Circumference of the coil =0.50 m
No. oftums =50.
A
Compute
Difficult
Trial Current through T.G. Deflections
No. inmA. 91 92 9) 94
1 500 rnA 40° 41° 41° 40°
2 800mA 45° 46° 46° 45°
c = 2 1t r = 0.500.5
2x3.142= 0.079 m. r=---
Trial No.1
IK =-- mean 9 = 41.5° (1 mark)
tanB'
236
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