15.Mechanical Effect of Electric Current (Physics - Chapter Wise Question Bank)

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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



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

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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



[[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



[[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'

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K <= 0.5 (1 mark)i tan 41.5"

)J n kBH = 0 (1 mark)

2r

Trial No.2 also to be done.

15.Mechanical Effect of Electric Current (Physics - Chapter Wise Question Bank)

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