CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two long parallel wires carrying currents 2.0 A and 4.0 A respectively in opposite directions. The separation between the wires is 0.25 m. Calculate the force per unit length between them in the order of 106.Also, state whether it will be attractive or repulsive.Take magnetic permeability as μ0=4π×107N/A2
Formula to be used, F(per unit length)=μ0I1I22πd


A

6.4, repulsive

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

4.5, attractive

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

6, attractive

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

4.5, repulsive

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

6.4, repulsive


Given
Current,I1=2.0A
Current,I2=4.0A
Distance,d =0.25 m
permeability,(μ0)=4π×107N/A2
Formula to be used, F=μ0I1I22πd
Substituting the given values, F=6.4×106N/m
Since the currents are in opposite directions, they are anti-parallel and hence have a repulsive force between them.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Force on a Current Carrying Conductor in a Magnetic Field
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon