The figure shows two infinitely long and this parallel wires P and Q placed 0 . 2 m apart perpendicular to the plane of the paper, carrying equal currents of 20 A each. Find the magnetic field at a point R , at a distance of 0 . 2 m from both the wires. The symbol ⊗ indicates that the current is directed inward.

Point R is situated equidistant from wires P and Q . Hence, for the given case the magnitude of the magnetic field at R due to both the wires will be same. That is, B P = B Q = μ ∘ I 2 π d = 2 × 10 - 7 H m - 1 20 A 0 . 2 m = 2 × 10 - 5 T Since, in both the wires current is directed inward then: Since, the two wires carry current in the same direction then the net magnetic field at point R : B R = 2 B cos 30 ° = 2 2 × 10 - 5 T 3 2 = 2 3 × 10 - 5 T Hence, the magnetic field at point R is 2 3 × 10 - 5 T .

The velocity vector of a proton is directed perpendicular to the plane of a current loop at its centre O as shown in the figure. What will be the force acting on the proton at the centre of the loop?

The magnetic force acting on the charged particle: F = q v × B The magnetic field due to the current carrying loop is along the axis of the current carrying loop, this is also the direction of the velocity of the proton. So, the angle between the velocity vector v and Magnetic field B is 0 ° . So, the magnetic force on the proton will be: F = q v B sin 0 ° = 0 N . Any charged particle moving towards the centre of the circular current carrying loop along the axis will not suffer any magnetic force due to the current carrying loop. Hence, in the given condition no force will be acting on the proton due to the current carrying loop.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Goa Board-2019>Q 5

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To convert a galvanometer into a voltmeter, we must connect a

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Important Questions on Moving Charges and Magnetism

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Goa Board-2019>Q 6

Figure shows a circular coil with centre ‘O’. If current flows through the coil, the magnetic field produced by it is maximum at :

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Goa Board-2019>Q 7

The figure shows four directions for the velocity vector v of a positively charged particle. It is moving through a uniform electric field E, directed normally out of the page and a uniform magnetic field B as shown :

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Which direction of motion will give the particle the greatest net force ?

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Goa Board-2019>Q 8

The figure shows four directions for the velocity vector v of a positively charged particle. It is moving through a uniform electric field E, directed normally out of the page and a uniform magnetic field B as shown :

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Motion in which direction may result in net force zero on the particle.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 4 - Moving Charges and Magnetism>Goa Board-2019>Q 9

The figure shows four directions for the velocity vector v of a positively charged particle. It is moving through a uniform electric field E, directed normally out of the page and a uniform magnetic field B as shown :

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