I have one question,that i can't find a solution to.So if someone can help,i ll be really grateful!

Q-A Circular loop of radius 20 cm carries a current 10 A.An electron crosses the plane of the loop with a speed of 2.0 x 10^6 m/s (2 into 10 to the power 6 meters/second).The direction of motion makes an angle of 30 degrees with the axis of the circle and passes through its centre. What is the magnitude of the magnetic force on the electron at the instant it crosses the plane?

 

Thanks in advance!

You need to find the magnetic field from the loop and then use the Lorentz force law to find the force. Do you know these equations?

You need to find the magnetic field from the loop and then use the Lorentz force law to find the force. Do you know these equations?

Magnetic field due to a circular loop,i know the equation to find Magnetic field on axial lines but here direction of motion makes an angle of 30 degs with the axis.If you know the equation for that please tell me.

And i know the equation for Lorentz force = q(V x B)+ qE.

Magnetic field due to a circular loop,i know the equation to find Magnetic field on axial lines but here direction of motion makes an angle of 30 degs with the axis.If you know the equation for that please tell me.

And i know the equation for Lorentz force = q(V x B)+ qE.

 

(qv X B) means it's a cross product, so the angle between the velocity vector and the magnetic field vector will be included there.